**Title: Understanding Hypothesis Testing in Movie-Going Habits****Introduction:**This educational content explores a hypothesis test concerning the average number of movies watched by moviegoers. A moviegoer is defined as someone who watches at least one movie in a theater over a 12-month period. We examine whether a sample mean differs significantly from a known national average.**Problem Statement:**The average moviegoer sees 8.5 movies per year. A sample of 42 moviegoers from a large university reported an average of 10.2 movies per year. Given a population standard deviation of 3.2 movies, we will perform a hypothesis test at the 0.01 level of significance to determine if this difference is significant.**Part 1 of 5: Stating the Hypotheses**1. **Null Hypothesis (H₀):** μ = 8.5; the average number of movies watched is 8.5.2. **Alternative Hypothesis (H₁):** μ ≠ 8.5; the average number of movies watched is not 8.5.This scenario requires a two-tailed test, which checks if the sample mean is significantly different, either higher or lower, than the population mean.**Part 2 of 5: Critical Values**- To find the critical value(s), calculations must be rounded to at least two decimal places. Should there be multiple critical values, they should be separated by commas.**Note on Graphs/Diagrams:**- This segment does not include specific graphs or diagrams. However, in a typical educational setting, Z-distribution tables or graphs illustrating the two-tailed hypothesis test would be helpful in visualizing the critical regions.This educational example guides students in understanding the process of hypothesis testing using real-world data and statistical reasoning.

Given Sample size n=42, sd=3.2, population mean μ=8.5, level of significance ɑ=0.01

Answered: Moviegoers The average "moviegoer" sees…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A marketing researcher wants to estimate the mean amount spent per year ($) on a web site by membership member shoppers. Suppose a random sample of 100 membership member shoppers who recently made a purchase on the web site yielded a mean amount spent of $55 and a standard deviation of $52. Complete parts (a) and (b) below. a. Is there evidence that the population mean amount spent per year on the web site by membership member shoppers is different from $51? (Use a 0.01 level of significance.) State the null and alternative hypotheses. Ho: H (Type integers or decimals. Do not round. Do not include the $ symbol in your answer.) Identify the critical value(s). The critical value(s) is/are (Type an integer or a decimal. Round to two decimal places as needed. Use a comma to separate answers as needed.) Determine the test statistic. The test statistic, 1STAT, is (Type an integer or a decimal. Round to two decimal places as needed.) State the conclusion. Ho. There is evidence that the…
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light examined data on employment and answered questions regarding why workers separate from their employes. According to the article, the standard deviation of the length of time that women with one job are employed during the first 8 years of their career is 92 weeks. Length of time employed during the first 8 years of career is a left skewed variable. For that variable, do the following tasks. A. determine the sampling distribution of the sample mean for simple random samples of 50 women with one job. Explain your reasoning B. Obtain the probability that the sampling error made in estimating the mean length of time employed by all women with one job by that of a random sample of 50 such women will be at most 20 weeks
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A parenting magazine reports that the average amount of wireless data used by teenagers each month is 5 Gb. For her science fair project, Ella sets out to prove the magazine wrong. She claims that the mean among teenagers in her area is less than reported. Ella collects information from a simple random sample of 25 teenagers at her high school, and calculates a mean of 4.7 Gb per month with a standard deviation of 0.9 Gb per month. Assume that the population distribution is approximately normal. Test Ella's claim at the 0.01 level of significance. Step 3 of 3: Draw a conclusion and interpret the decision. E Tables E Keypad Answer Keyboard Shortcuts We reject the null hypothesis and conclude that there is sufficient evidence at a 0.01 level of significance that the average amount of wireless data used by teenagers each month is less than 5 Gb. We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.01 level of significance that the average amount of…

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