What is the optimal time for a scuba diver to be on the bottom of the ocean? That depends on the depth of the dive. The U.S. Navy has done a lot of research on this topic. The Navy defines the "optimal time" to be the time at each depth for the best balance between length of work period and decompression time after surfacing. Let displaystyle{x}= depth of dive in meters, and let displaystyle{y}= optimal time in hours. A random sample of divers gave the following data. begin{array}{|c|c|} hline x & 13.1 & 23.3 & 31.2 & 38.3 & 51.3 &20.5 & 22.7 hline y & 2.78 & 2.18 & 1.48 & 1.03 & 0.75 & 2.38 & 2.20 hline end{array} (a) Find displaystyleΣ{x},Σ{y},Σ{x}^{2},Σ{y}^{2},Σ{x}{y},{quadtext{and}quad}{r}. (Round r to three decimal places.) displaystyleΣ{x}= displaystyleΣ{y}= displaystyleΣ{x}^

Question
Confidence intervals
asked 2021-02-12
What is the optimal time for a scuba diver to be on the bottom of the ocean? That depends on the depth of the dive. The U.S. Navy has done a lot of research on this topic. The Navy defines the "optimal time" to be the time at each depth for the best balance between length of work period and decompression time after surfacing. Let \(\displaystyle{x}=\) depth of dive in meters, and let \(\displaystyle{y}=\) optimal time in hours. A random sample of divers gave the following data.
\(\begin{array}{|c|c|} \hline x & 13.1 & 23.3 & 31.2 & 38.3 & 51.3 &20.5 & 22.7 \\ \hline y & 2.78 & 2.18 & 1.48 & 1.03 & 0.75 & 2.38 & 2.20 \\ \hline \end{array}\)
(a)
Find \(\displaystyleΣ{x},Σ{y},Σ{x}^{2},Σ{y}^{2},Σ{x}{y},{\quad\text{and}\quad}{r}\). (Round r to three decimal places.)
\(\displaystyleΣ{x}=\)
\(\displaystyleΣ{y}=\)
\(\displaystyleΣ{x}^{2}=\)
\(\displaystyleΣ{y}^{2}=\)
\(\displaystyleΣ{x}{y}=\)
\(\displaystyle{r}=\)
(b)
Use a \(1\%\) level of significance to test the claim that \(\displaystyle\rho<{0}\)</span>. (Round your answers to two decimal places.)
\(\displaystyle{t}=\)
critical \(\displaystyle{t}=\)
Conclusion
Reject the null hypothesis. There is sufficient evidence that \(\displaystyle\rho<{0}\)</span>.Reject the null hypothesis. There is insufficient evidence that \(\displaystyle\rho<{0}\)</span>.
Fail to reject the null hypothesis. There is sufficient evidence that \(\displaystyle\rho<{0}\)</span>.Fail to reject the null hypothesis. There is insufficient evidence that \(\displaystyle\rho<{0}.\)</span>
(c)
Find \(\displaystyle{S}_{{e}},{a},{\quad\text{and}\quad}{b}\). (Round your answers to four decimal places.)
\(\displaystyle{S}_{{e}}=\)
\(\displaystyle{a}=\)
\(\displaystyle{b}=\)

Answers (1)

2021-02-13
Step 1
(a)
The value of \(\displaystyle\sum{x}\) is,
\(\displaystyle\sum{x}={13.1}+{23.3}+{31.2}+{38.5}+{51.3}+{20.5}+{22.7}={200.6}\)
Thus, the value of \(\displaystyle\sum{x}\ {i}{s}\ {200.6}\)
The value of \(\displaystyle\sum{y}\) is,
\(\displaystyle\sum{y}={2.78}+{2.18}+{1.48}+{1.03}+{0.75}+{2.38}+{2.2}={12.8}\)
Thus, the value of \(\displaystyle\sum{y}\ {i}{s}\ {12.8}\)
The value of \(\displaystyle\sum{x}^{2}\) is,
\(\displaystyle\sum{x}^{2}={13.1}^{2}+{23.3}^{2}+{31.2}^{2}+{38.5}^{2}+{51.3}^{2}+{20.5}^{2}+{22.7}^{2}\)
\(\displaystyle={171.61}+{542.89}+{973.44}+{1482.25}+{2631.69}+{420.25}+{515.29}\)
\(\displaystyle={6737.42}\) Thus, the value of \(\displaystyle\sum{x}^{2}\ {i}{s}\ {6737.42}\)
Thus, the value of \(\displaystyle\sum{y}^{2}\) is,
\(\displaystyle\sum{y}^{2}={2.78}^{2}+{2.18}^{2}+{1.48}^{2}+{1.03}^{2}+{0.75}^{2}+{2.38}^{2}+{2.2}^{2}\)
\(\displaystyle={7.7284}+{4.7524}+{2.1904}+{1.0609}+{0.5625}+{5.6644}+{4.84}\)
\(\displaystyle={26.7990}\)
Thus, the value of \(\displaystyle\sum{y}^{2}\ {i}{s}\ {26.7990}.\)
The value of \(\displaystyle\sum{x}{y}\) is,
\(\displaystyle\sum{x}{y}={\left({13.1}\times{2.78}\right)}+{\left({23.3}\times{2.18}\right)}+{\left({31.2}\times{1.48}\right)}+{\left({38.5}\times{1.03}\right)}+{\left({51.3}\times{0.75}\right)}+{\left({20.5}\times{2.38}\right)}+{\left({22.7}\times{2.2}\right)}\)
\(\displaystyle={36.418}+{50.794}+{46.176}+{39.655}+{38.475}+{48.790}+{49.940}\)
\(\displaystyle={310.248}\)
Thus, the value of \(\displaystyle\sum{x}{y}\ {i}{s}\ {310.248}.\)
The value of r is,
\(\displaystyle{r}=\frac{{{n}\sum{x}{y}-{\left(\sum{x}\right)}{\left(\sum{y}\right)}}}{\sqrt{{{\left[{n}\sum{x}^{2}-{\left(\sum{x}\right)}^{2}\right]}{\left[\sum{y}^{2}-{\left(\sum{y}\right)}^{2}\right]}}}}\)
\(\displaystyle=\frac{{{7}{\left({310.248}\right)}-{\left({200.6}\times{12.8}\right)}}}{{\sqrt{{{\left[{\left({7}\times{6737.42}\right)}-{\left({200.6}\right)}^{2}\right]}{\left[{\left({7}\times{26.7990}\right)}-{\left({12.8}\right)}^{2}\right]}}}}}\)
\(\displaystyle=\frac{{-{395.944}}}{{405.4729}}\)
\(\displaystyle=-{0.976}\)
Hence, the correlation coefficient r is \(\displaystyle-{0.976}.\)
Step 2
(b)
The hypothesis is,
Null hypothesis:
\(\displaystyle{H}_{{0}}:\rho={0}\)
Alternative hypothesis:
\(\displaystyle{H}_{{1}}:\rho<{0}\)</span>
The test statistic is,
\(\displaystyle{t}=-{r}\sqrt{{\frac{{{n}-{2}}}{{{1}-{r}^{2}}}}}\)
\(\displaystyle=-{0.976}\sqrt{{\frac{{{7}-{2}}}{{{1}-{\left(-{0.976}\right)}^{2}}}}}\)
\(\displaystyle=\frac{{-{2.1824}}}{{0.2178}}\)
\(\displaystyle=-{10.02}\)
Thus, the test statistic is \(\displaystyle-{10.02}.\)
The degrees of freedom is,
\(\displaystyle{d}{f}={n}-{2}\)
\(\displaystyle={7}-{2}\)
\(\displaystyle={5}\)
The degrees of freedom is 5.
Decision rules for correlation coefficient:
If the test statistic is less than the negative critical value, then reject the null hypothesis.
Computation of critical value:
The critical value of t-distribution for 0.01 level of significance at 5 degrees of freedom can be obtained using the excel formula “=T.INV(0.01,5)”. The critical value is \(\displaystyle-{3.36}.\)
Thus, the critical value is \(\displaystyle-{3.36}.\)
Conclusion:
The test statistic is \(\displaystyle-{10.02}\) less than critical value \(\displaystyle-{3.36}.\)
Based on the decision rule, reject the null hypothesis.
Reject the null hypothesis. There is sufficient evidence that \(\displaystyle\rho<{0}.\)</span>
Step 3
c)
The value of \(\displaystyle\overline{{x}}\) is,
\(\displaystyle\overline{{x}}=\frac{{\sum{x}}}{{n}}\)
\(\displaystyle=\frac{200.6}{{7}}\)
\(\displaystyle={28.6571}\)
The value of \(\displaystyle\overline{{y}}\) is,
\(\displaystyle\overline{{y}}=\frac{{\sum{y}}}{{n}}\)
\(\displaystyle=\frac{12.8}{{7}}\)
\(\displaystyle={1.8286}\)
The value of b is,
\(\displaystyle{b}=\frac{{{n}\sum{x}{y}-{\left(\sum{x}\right)}{\left(\sum{y}\right)}}}{{{n}\sum{x}^{2}-{\left(\sum{x}\right)}^{2}}}\)
\(\displaystyle=\frac{{{7}{\left({310.248}\right)}-{\left({200.6}\times{12.8}\right)}}}{{{\left({7}\times{6737.42}\right)}-{\left({200.6}\right)}^{2}}}\)
\(\displaystyle=-\frac{395.944}{{6921.58}}\)
\(\displaystyle=-{0.0572}\)
Thus, the value of b is \(\displaystyle-{0.0572}.\)
The value of a is,
\(\displaystyle{a}=\overline{{y}}-{b}\overline{{x}}\)
\(\displaystyle={1.8286}-{\left(-{0.0572}\times{28.6571}\right)}\)
\(\displaystyle={1.8286}+{1.6392}\)
\(\displaystyle={3.4678}\)
Thus, the value of a is 3.4678.
The value of \(\displaystyle{S}_{{e}}\) is,
\(\displaystyle{S}_{{e}}=\sqrt{{\frac{{\sum{y}^{2}-{a}\sum{y}-{b}\sum{x}{y}}}{{{n}-{2}}}}}\)
\(\displaystyle=\sqrt{{\frac{{{26.7990}-{\left({3.4678}\times{12.8}\right)}-{\left(-{0.0572}\times{310.248}\right)}}}{{{7}-{2}}}}}\)
\(\displaystyle=\sqrt{{\frac{{{0.1573456}}}{{5}}}}\)
\(\displaystyle={0.1774}\)
Hence, the value of \(\displaystyle{S}_{{e}}\) is 0.1776.
0

Relevant Questions

asked 2021-01-17
A new thermostat has been engineered for the frozen food cases in large supermarkets. Both the old and new thermostats hold temperatures at an average of \(25^{\circ}F\). However, it is hoped that the new thermostat might be more dependable in the sense that it will hold temperatures closer to \(25^{\circ}F\). One frozen food case was equipped with the new thermostat, and a random sample of 21 temperature readings gave a sample variance of 5.1. Another similar frozen food case was equipped with the old thermostat, and a random sample of 19 temperature readings gave a sample variance of 12.8. Test the claim that the population variance of the old thermostat temperature readings is larger than that for the new thermostat. Use a \(5\%\) level of significance. How could your test conclusion relate to the question regarding the dependability of the temperature readings? (Let population 1 refer to data from the old thermostat.)
(a) What is the level of significance?
State the null and alternate hypotheses.
\(H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}>?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}\neq?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}?_{2}^{2},H1:?_{1}^{2}=?_{2}^{2}\)
(b) Find the value of the sample F statistic. (Round your answer to two decimal places.)
What are the degrees of freedom?
\(df_{N} = ?\)
\(df_{D} = ?\)
What assumptions are you making about the original distribution?
The populations follow independent normal distributions. We have random samples from each population.The populations follow dependent normal distributions. We have random samples from each population.The populations follow independent normal distributions.The populations follow independent chi-square distributions. We have random samples from each population.
(c) Find or estimate the P-value of the sample test statistic. (Round your answer to four decimal places.)
(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?
At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the ? = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.
(e) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings.Fail to reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings. Fail to reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.Reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.
asked 2020-12-07
Would you rather spend more federal taxes on art? Of a random sample of \(n_{1} = 86\) politically conservative voters, \(r_{1} = 18\) responded yes. Another random sample of \(n_{2} = 85\) politically moderate voters showed that \(r_{2} = 21\) responded yes. Does this information indicate that the population proportion of conservative voters inclined to spend more federal tax money on funding the arts is less than the proportion of moderate voters so inclined? Use \(\alpha = 0.05.\) (a) State the null and alternate hypotheses. \(H_0:p_{1} = p_{2}, H_{1}:p_{1} > p_2\)
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asked 2020-10-23
A random sample of \(\displaystyle{n}_{{1}}={16}\) communities in western Kansas gave the following information for people under 25 years of age.
\(\displaystyle{X}_{{1}}:\) Rate of hay fever per 1000 population for people under 25
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\(\displaystyle{X}_{{2}}:\) Rate of hay fever per 1000 population for people over 50
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(ii) Assume that the hay fever rate in each age group has an approximately normal distribution. Do the data indicate that the age group over 50 has a lower rate of hay fever? Use \(\displaystyle\alpha={0.05}.\)
(a) What is the level of significance?
State the null and alternate hypotheses.
\(\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}<\mu_{{2}}\)
\(\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}>\mu_{{2}}\)
\(\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}\ne\mu_{{2}}\)
\(\displaystyle{H}_{{0}}:\mu_{{1}}>\mu_{{2}},{H}_{{1}}:\mu_{{1}}=\mu_{{12}}\)
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The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations,
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations,
The Student's t. We assume that both population distributions are approximately normal with known standard deviations,
What is the value of the sample test statistic? (Test the difference \(\displaystyle\mu_{{1}}-\mu_{{2}}\). Round your answer to three decimalplaces.)
What is the value of the sample test statistic? (Test the difference \(\displaystyle\mu_{{1}}-\mu_{{2}}\). Round your answer to three decimal places.)
(c) Find (or estimate) the P-value.
P-value \(\displaystyle>{0.250}\)
\(\displaystyle{0.125}<{P}-\text{value}<{0},{250}\)
\(\displaystyle{0},{050}<{P}-\text{value}<{0},{125}\)
\(\displaystyle{0},{025}<{P}-\text{value}<{0},{050}\)
\(\displaystyle{0},{005}<{P}-\text{value}<{0},{025}\)
P-value \(\displaystyle<{0.005}\)
Sketch the sampling distribution and show the area corresponding to the P-value.
P.vaiue Pevgiue
P-value f P-value
asked 2020-11-08
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asked 2020-11-08
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asked 2021-02-09
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Using the Minitab statistical analysis program to enter the data and perform the analysis, complete the following: Construct a one-sided \(\displaystyle{95}\%\) confidence interval for the true difference in population means. Test the null hypothesis that the population means are identical at the 0.05 level of significance.
asked 2021-03-06
Use either the critical-value approach or the P-value approach to perform the required hypothesis test. Approximately 450,000 vasectomies are performed each year in the United States. In this surgical procedure for contraception, the tube carrying sperm from the testicles is cut and tied. Several studies have been conducted to analyze the relationship between vasectomies and prostate cancer. The results of one such study by E. Giovannucci et al. appeared in the paper “A Retrospective Cohort Study of Vasectomy and Prostate Cancer in U.S. Men” (Journal of the American Medical Association, Vol. 269(7), pp. 878-882). Of 21,300 men who had not had a vasectomy, 69 were found to have prostate cancer, of 22,000 men who had had a vasectomy, 113 were found to have prostate cancer. a. At the 1% significance level, do the data provide sufficient evidence to conclude that men who have had a vasectomy are at greater risk of having prostate cancer? b. Is this study a designed experiment or an observational study? Explain your answer. c. In view of your answers to parts (a) and (b), could you reasonably conclude that having a vasectomy causes an increased risk of prostate cancer? Explain your answer.
asked 2021-02-25
Iron is very important for babies' growth. A common belief is that breastfeeding will help the baby to get more iron than formula feeding. To justify the belief, a study followed 2 groups of babies from born to 6 months. With one group babies are breast fed, and the other group are formula fed without iron supplements. Data below shows iron levels of those two groups of babies. \(\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}{G}{r}{o}{u}{p}&{S}{a}\mp\le\ {s}{i}{z}{e}&{m}{e}{a}{n}&{S}{\tan{{d}}}{a}{r}{d}\ {d}{e}{v}{i}{a}{t}{i}{o}{n}\backslash{h}{l}\in{e}{B}{r}{e}\ast-{f}{e}{d}&{23}&{13.3}&{1.7}\backslash{h}{l}\in{e}{F}{\quad\text{or}\quad}\mu{l}{a}-{f}{e}{d}&{23}&{12.4}&{1.8}\backslash{h}{l}\in{e}{D}{I}{F}{F}={B}{r}{e}\ast-{F}{\quad\text{or}\quad}\mu{l}{a}&{23}&{0.9}&{1.4}\backslash{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\) (1) There are two groups we need to compare for the study: Breast-Fed and Formula- Fed. Are those two groups dependent or independent? Based on your answer, what inference procedure should we apply for this research? (2) Please perform the inference you decided in (1), and make sure to follow the 5-step procedure for any hypothesis test. (3) Based on your conclusion in (2), what kind of error could you make? Explain the type of error using the context words for this research
asked 2021-03-11
An automobile tire manufacturer collected the data in the table relating tire pressure x​ (in pounds per square​ inch) and mileage​ (in thousands of​ miles). A mathematical model for the data is given by
\(\displaystyle​ f{{\left({x}\right)}}=-{0.554}{x}^{2}+{35.5}{x}-{514}.\)
\(\begin{array}{|c|c|} \hline x & Mileage \\ \hline 28 & 45 \\ \hline 30 & 51\\ \hline 32 & 56\\ \hline 34 & 50\\ \hline 36 & 46\\ \hline \end{array}\)
​(A) Complete the table below.
\(\begin{array}{|c|c|} \hline x & Mileage & f(x) \\ \hline 28 & 45 \\ \hline 30 & 51\\ \hline 32 & 56\\ \hline 34 & 50\\ \hline 36 & 46\\ \hline \end{array}\)
​(Round to one decimal place as​ needed.)
\(A. 20602060xf(x)\)
A coordinate system has a horizontal x-axis labeled from 20 to 60 in increments of 2 and a vertical y-axis labeled from 20 to 60 in increments of 2. Data points are plotted at (28,45), (30,51), (32,56), (34,50), and (36,46). A parabola opens downward and passes through the points (28,45.7), (30,52.4), (32,54.7), (34,52.6), and (36,46.0). All points are approximate.
\(B. 20602060xf(x)\)
Acoordinate system has a horizontal x-axis labeled from 20 to 60 in increments of 2 and a vertical y-axis labeled from 20 to 60 in increments of 2.
Data points are plotted at (43,30), (45,36), (47,41), (49,35), and (51,31). A parabola opens downward and passes through the points (43,30.7), (45,37.4), (47,39.7), (49,37.6), and (51,31). All points are approximate.
\(C. 20602060xf(x)\)
A coordinate system has a horizontal x-axis labeled from 20 to 60 in increments of 2 and a vertical y-axis labeled from 20 to 60 in increments of 2. Data points are plotted at (43,45), (45,51), (47,56), (49,50), and (51,46). A parabola opens downward and passes through the points (43,45.7), (45,52.4), (47,54.7), (49,52.6), and (51,46.0). All points are approximate.
\(D.20602060xf(x)\)
A coordinate system has a horizontal x-axis labeled from 20 to 60 in increments of 2 and a vertical y-axis labeled from 20 to 60 in increments of 2. Data points are plotted at (28,30), (30,36), (32,41), (34,35), and (36,31). A parabola opens downward and passes through the points (28,30.7), (30,37.4), (32,39.7), (34,37.6), and (36,31). All points are approximate.
​(C) Use the modeling function​ f(x) to estimate the mileage for a tire pressure of 29
\(\displaystyle​\frac{{{l}{b}{s}}}{{{s}{q}}}\in.\) and for 35
\(\displaystyle​\frac{{{l}{b}{s}}}{{{s}{q}}}\in.\)
The mileage for the tire pressure \(\displaystyle{29}\frac{{{l}{b}{s}}}{{{s}{q}}}\in.\) is
The mileage for the tire pressure \(\displaystyle{35}\frac{{{l}{b}{s}}}{{{s}{q}}}\) in. is
(Round to two decimal places as​ needed.)
(D) Write a brief description of the relationship between tire pressure and mileage.
A. As tire pressure​ increases, mileage decreases to a minimum at a certain tire​ pressure, then begins to increase.
B. As tire pressure​ increases, mileage decreases.
C. As tire pressure​ increases, mileage increases to a maximum at a certain tire​ pressure, then begins to decrease.
D. As tire pressure​ increases, mileage increases.
asked 2020-10-23
1. Find each of the requested values for a population with a mean of \(? = 40\), and a standard deviation of \(? = 8\) A. What is the z-score corresponding to \(X = 52?\) B. What is the X value corresponding to \(z = - 0.50?\) C. If all of the scores in the population are transformed into z-scores, what will be the values for the mean and standard deviation for the complete set of z-scores? D. What is the z-score corresponding to a sample mean of \(M=42\) for a sample of \(n = 4\) scores? E. What is the z-scores corresponding to a sample mean of \(M= 42\) for a sample of \(n = 6\) scores? 2. True or false: a. All normal distributions are symmetrical b. All normal distributions have a mean of 1.0 c. All normal distributions have a standard deviation of 1.0 d. The total area under the curve of all normal distributions is equal to 1 3. Interpret the location, direction, and distance (near or far) of the following zscores: \(a. -2.00 b. 1.25 c. 3.50 d. -0.34\) 4. You are part of a trivia team and have tracked your team’s performance since you started playing, so you know that your scores are normally distributed with \(\mu = 78\) and \(\sigma = 12\). Recently, a new person joined the team, and you think the scores have gotten better. Use hypothesis testing to see if the average score has improved based on the following 8 weeks’ worth of score data: \(82, 74, 62, 68, 79, 94, 90, 81, 80\). 5. You get hired as a server at a local restaurant, and the manager tells you that servers’ tips are $42 on average but vary about \($12 (\mu = 42, \sigma = 12)\). You decide to track your tips to see if you make a different amount, but because this is your first job as a server, you don’t know if you will make more or less in tips. After working 16 shifts, you find that your average nightly amount is $44.50 from tips. Test for a difference between this value and the population mean at the \(\alpha = 0.05\) level of significance.
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