Construct a 95% confidence interval for the true average monthly salary earned by all employees of People Plus Pty, if it was found that the average monthly salary earned by a sample of 19 employees of the company was R18 500, with a standard deviation of R1 750. Interpret your answer

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Construct a 95% confidence interval for the true average monthly salary earned by all employees of People Plus Pty, if it was found that the average monthly salary earned by a sample of 19 employees of the company was R18 500, with a standard deviation of R1 750. Interpret your answer

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asked 2021-05-05

A random sample of $ n_1 = 14 $ winter days in Denver gave a sample mean pollution index $ x_1 = 43 $.
Previous studies show that $ \sigma_1 = 19 $.
For Englewood (a suburb of Denver), a random sample of $ n_2 = 12 $ winter days gave a sample mean pollution index of $ x_2 = 37 $.
Previous studies show that $ \sigma_2 = 13 $.
Assume the pollution index is normally distributed in both Englewood and Denver.
(a) State the null and alternate hypotheses.
$ H_0:\mu_1=\mu_2.\mu_1>\mu_2 $
$ H_0:\mu_1<\mu_2.\mu_1=\mu_2 $
$ H_0:\mu_1=\mu_2.\mu_1<\mu_2 $
$ H_0:\mu_1=\mu_2.\mu_1\neq\mu_2 $
(b) What sampling distribution will you use? What assumptions are you making? NKS The Student's t. We assume that both population distributions are approximately normal with known standard deviations.
The standard normal. 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 known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.
(c) What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate.
(Test the difference $ \mu_1 - \mu_2 $. Round your answer to two decimal places.) NKS (d) Find (or estimate) the P-value. (Round your answer to four decimal places.)
(e) Based on your answers in parts (i)−(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \alpha?
At the $ \alpha = 0.01 $ level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the $ \alpha = 0.01 $ level, we reject the null hypothesis and conclude the data are statistically significant.
At the $ \alpha = 0.01 $ level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the $ \alpha = 0.01 $ level, we reject the null hypothesis and conclude the data are not statistically significant.
(f) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver. (g) Find a 99% confidence interval for
$ \mu_1 - \mu_2 $.
(Round your answers to two decimal places.)
lower limit
upper limit
(h) Explain the meaning of the confidence interval in the context of the problem.
Because the interval contains only positive numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, we can not say that the mean population pollution index for Englewood is different than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains only negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is less than that of Denver.

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asked 2021-03-06

A sample of 51 research cotton samples resulted in a sample average percentage elongation of 8.19 and a sample standard deviation of 1.45. Calculate a $95\%$ large-sample CI for the true average percentage elongation $\mu$ (Round your answer to three decimal places.)

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asked 2021-02-23

1. A researcher is interested in finding a 98% confidence interval for the mean number of times per day that college students text. The study included 144 students who averaged 44.7 texts per day. The standard deviation was 16.5 texts. a. To compute the confidence interval use a ? z t distribution. b. With 98% confidence the population mean number of texts per day is between and texts. c. If many groups of 144 randomly selected members are studied, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population number of texts per day and about percent will not contain the true population mean number of texts per day. 2. You want to obtain a sample to estimate how much parents spend on their kids birthday parties. Based on previous study, you believe the population standard deviation is approximately $\displaystyle\sigma={40.4}$ dollars. You would like to be 90% confident that your estimate is within 1.5 dollar(s) of average spending on the birthday parties. How many parents do you have to sample? n = 3. You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately $\displaystyle\sigma={57.5}$. You would like to be 95% confident that your estimate is within 0.1 of the true population mean. How large of a sample size is required?

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asked 2021-03-11

The local energy company claims the average annual electricity bill for its subscribers is just $600. A consumer watchdog group wants to dispute this claim. All agree that the standard deviation sigma of annual electricity bills is $150.
Some time later, a wealthy activist provides funding for a simple random sample of 250 households. The average annual electricity bill for this sample is $622. Find a $95\%$ confidence interval for the true mean annual electric bill, based on this sample.

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asked 2020-11-12

(a) The company's production equipment produces metal discs weighing 200 g. It should be noted that the weight of the discs corresponds to the normal distribution. To check machine consistency, 20 discs are randomly selected with an average weight of 205 g and a standard deviation of 7 g. What is the 99% confidence interval for the average weight of the selected discs?
(b) A company launched a new model of golf ball. It claimed that the driving distance is at least 300m. A sample of 20 balls yields a sample mean of 295m and sample standard deviation of 8m. It is assumed that the driving distance is normally distributed.
Conduct an appropriate hypothesis testing at the 0.05 level of significance. Is there any evidence that the average travel distance stated by the company is true?

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asked 2021-03-06

Based on a sample of 80 recent Masters graduates (40 male and 40 female), the following information was made available regarding their annual salaries. The standard deviation of salaries for the male graduates was $40,000 and that for the female graduates was $25,000.
a) For the male graduates, what is the probability of obtaining a sample mean salary within $10,000 of the population mean?
b) Consider the same question in (a) but for the female graduates. In which case, males or females, do we have a higher probability of obtaining a sample estimate within $10,000 of the population mean? Why?
c) Suppose that the sample mean salary of females is $125,000. Develop a $95\%$ confidence interval estimate for the mean salary of all female graduates

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asked 2021-02-16

A poll in 2017 reported that 699 out of 1027 adults in a certain country believe that marijuana should be legalized. When this poll about the subject was first conducted in 1969, only 12% of rhe country supported legalizztion. Assume the conditions for using the CLT are met.
a) Find and interpet a 99% confidence interval for the proportion of adults in the country 2017 that believe marijuana should be legalized is (0.643, 0.718)
b) Find and interpret a 90%confidence interval for this population parameter. The 90% confidence interval for the proportion of adults in the country 2017 that believe marijuana should be legalized is (0.657, 0.705)
c)Find the margin of error for each of the confidence intervals found The margin of error of the 99% confidence interval is 0.039 and the margin of error of the 90% confidence interval is 0.025
d) Without computing it, how would the margin of error of an 80% confidence interval compare with the margin of error for 90% and 99% intervals? Construct the 80% confidence interval to see if your production was correct
How would a 80% interval compare with the others in the margin of error?

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asked 2021-02-06

You may need to use the appropriate appendix table or technology to answer this question.
Money reports that the average annual cost of the first year of owning and caring for a large dog in 2017 is $1,448. The Irish Red and White Setter Association of America has requested a study to estimate the annual first-year cost for owners of this breed. A sample of 50 will be used. Based on past studies, the population standard deviation is assumed known with $\displaystyle\sigma=\${230}.$
$\begin{matrix} 1,902 & 2,042 & 1,936 & 1,817 & 1,504 & 1,572 & 1,532 & 1,907 & 1,882 & 2,153 \\ 1,945 & 1,335 & 2,006 & 1,516 & 1,839 & 1,739 & 1,456 & 1,958 & 1,934 & 2,094 \\ 1,739 & 1,434 & 1,667 & 1,679 & 1,736 & 1,670 & 1,770 & 2,052 & 1,379 & 1,939\\ 1,854 & 1,913 & 2,163 & 1,737 & 1,888 & 1,737 & 2,230 & 2,131 & 1,813 & 2,118\\ 1,978 & 2,166 & 1,482 & 1,700 & 1,679 & 2,060 & 1,683 & 1,850 & 2,232 & 2,294 \end{matrix}$
(a)
What is the margin of error for a $95\%$ confidence interval of the mean cost in dollars of the first year of owning and caring for this breed? (Round your answer to nearest cent.)
(b)
The DATAfile Setters contains data collected from fifty owners of Irish Setters on the cost of the first year of owning and caring for their dogs. Use this data set to compute the sample mean. Using this sample, what is the $95\%$ confidence interval for the mean cost in dollars of the first year of owning and caring for an Irish Red and White Setter? (Round your answers to nearest cent.)
$_______ to $________

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asked 2020-10-28

A manager at ACME Equipment Sale and Rental wondered how offering a free two-year service warranty on its tractors might influence sales. For the next 500 customers who expressed interest in buying a tractor, 250 were randomly offered a warranty, and the rest did not receive. Ninety-three of those offering a warranty, and fifty-four of those not offering a warranty ended up buying a tractor.
a. Construct a $95\%$ confidence interval for the difference between the proportions of customers purchasing tractors with and without warranties. Be sure to check all necessary assumptions and interpret the interval.
b. Test the hypothesis that offering the warranty increases the proportion of customers who eventually purchase a tractor. Be sure to check all necessary assumptions, state the null and alternative hypotheses, obtain the p-value, and state your conclusion. Should a manager offer a warranty based on this test?

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asked 2020-12-29

The presidential election is coming. Five survey companies (A, B, C, D, and E) are doing survey to forecast whether or not the Republican candidate will win the election. Each company randomly selects a sample size between 1000 and 1500 people. All of these five companies interview people over the phone during Tuesday and Wednesday. The interviewee will be asked if he or she is 18 years old or above and U.S. citizen who are registered to vote. If yes, the interviewee will be further asked: will you vote for the Republican candidate? On Thursday morning, these five companies announce their survey sample and results at the same time on the newspapers. The results show that a% (from A), b% (from B), c% (from C), d% (from D), and e% (from E) will support the Republican candidate. The margin of error is plus/minus 3% for all results. Suppose that $\displaystyle{c}{>}{a}{>}{d}{>}{e}{>}{b}$. When you see these results from the newspapers, can you exactly identify which result(s) is (are) not reliable and not accurate? That is, can you identify which estimation interval(s) does (do) not include the true population proportion? If you can, explain why you can, if no, explain why you cannot and what information you need to identify. Discuss and explain your reasons. You must provide your statistical analysis and reasons.

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Answer 1

Step 1 The sample size is 19, mean is R18500, standard deviation is R1750. The degrees of freedom is, $\displaystyle{d}{f}={n}-{1}$
$\displaystyle={19}-{1}$
$\displaystyle={18}$ The degrees of freedom is 18. The confidence level is 95%, than the level of significance is 0.05. Computation of critical value: The critical value of t-distribution for 0.05 level of significance at 18 degrees of freedom can be obtained using the excel formula “=T.INV.2T(0.05,18)”. The critical value is 2.1009. Step 2 The 95% confidence interval for the true average monthly salary earned by all employees of People Plus Pty is, $\displaystyle{C}{I}=\overline{{{x}}}\pm{t}_{{{\frac{{\alpha}}{{{2}}}}}}{\left({\frac{{{s}}}{{\sqrt{{{n}}}}}}\right)}$
$\displaystyle={18500}\pm{2.1009}{\left({\frac{{{1750}}}{{\sqrt{{{19}}}}}}\right)}$
$\displaystyle={18500}\pm{843.464}$
$\displaystyle={\left({18500}-{843.464},{18500}+{843.464}\right)}$
$\displaystyle={\left({17656.536},{19343.464}\right)}$
$\displaystyle\approx{\left({17657},{19343}\right)}$ Thus, the 95% confidence interval for the true average monthly salary earned by all employees of People Plus Pty is (17657,19343). Interpretation: When repeated samples of size 19 are taken, about 95% of those confidence intervals would contain the true average monthly salary earned by all employees of People Plus Pty that lies between 17657 and 19343. That is, there is 95% confidence that the true average monthly salary earned by all employees of People Plus Pty that lies between 17657 and 19343.

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Construct a 95% confidence interval for the true average monthly salary earned by all employees of People Plus Pty, if it was found that the average monthly salary earned by a sample of 19 employees of the company was R18 500, with a standard deviation of R1 750. Interpret your answer

Construct a 95% confidence interval for the true average monthly salary earned by all employees of People Plus Pty, if it was found that the average monthly salary earned by a sample of 19 employees of the company was R18 500, with a standard deviation of R1 750. Interpret your answer

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