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 bir

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 bir

Question
Confidence intervals
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?

Answers (1)

2021-02-24
Step 1 Hello. Since you have posted multiple questions and not specified which question needs to be solved, we will solve the first question for you. If you want any other specific question to be solved, then please resubmit only that question or specify the question number. 1. a. The confidence level is 98%. Therefore, the level of significance is 1 – confidence level = 2% or 0.02. The sample size, n is provided as 144 whose average is 44.7. The standard deviation is 16.5. The data is continuous and the sample is 144 which is greater than 30. Therefore, it is assumed that the data follows the normal distribution making the z distribution appropriate to obtain the confidence interval. Step 2 b. The 98% confidence interval for population mean number of texts per day is, \(\displaystyle{C}.{I}.=\overline{{{x}}}\pm{z}_{{\frac{{\alpha}}{{{2}}}}}{\left\lbrace{\frac{{{s}}}{{\sqrt{{{n}}}}}}\right\rbrace}\)
\(\displaystyle={44.7}\pm{z}_{{\frac{{{0.02}}}{{{2}}}}}{\frac{{{16.5}}}{{\sqrt{{{144}}}}}}\)
\(\displaystyle={44.7}\pm{z}_{{{0.01}}}{\frac{{{16.5}}}{{\sqrt{{{144}}}}}}\)
\(\displaystyle{44.7}\pm{\left({2.33}\right)}{\frac{{{16.5}}}{{\sqrt{{{144}}}}}}\) (Using standard normal table) \(\displaystyle={\left({41.496},{47.904}\right)}\) c. If many groups of 144 randomly selected members are studied, then a different confidence interval would be produced from each group. About 98% of these confidence intervals will contain the true population number of texts per day and about 2% (100 – 98) will not contain the true population mean number of texts per day.
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