**Educational Website Content: Hypothesis Testing Problem**Title: Analyzing Web Browser Market Share at a University---**Problem Statement:**The worldwide market share for a particular web browser was 20.1% in a recent month. Suppose that a sample of 120 random students at a certain university finds that 30 students use this browser. We aim to evaluate whether the market share for this web browser at the university is greater than the worldwide market share of 20.1%.---**Task (Part a):****At the 0.01 level of significance, is there evidence that the market share for the web browser at the university is greater than the worldwide market share of 20.1%?****Steps to Solve:**1. **Determine the Null and Alternative Hypotheses:** - **Option A:** $ H_0: \pi \neq 0.201 $; $ H_1: \pi < 0.201 $ - **Option B:** $ H_0: \pi = 0.201 $; $ H_1: \pi \neq 0.201 $ - **Correct Option (C):** $ H_0: \pi \le 0.201 $; $ H_1: \pi > 0.201 $ - **Option D:** $ H_0: \pi \neq 0.201 $; $ H_1: \pi = 0.201 $2. **Calculate the Test Statistic:** Formula for calculating the test statistic $ Z_{\text{STAT}}: $ \[ Z_{\text{STAT}} = \, \text{(Type an integer or a decimal. Round to two decimal places as needed.)} \] ---Note: Ensure your statistical calculations, steps, hypothesis setting, and decision-making process are correct and properly rounded when using this information in practical applications.

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Answered: The worldwide market share for a web…

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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You are conducting a study to see if the proportion of voters who prefer the Democratic candidate is significantly different from 54% at a level of significance of a = 0.05. According to your sample, 33 out of 56 potential voters prefer the Democratic candidate. a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: Ho: ? Select an answer ✓ (please enter a decimal) H₁: ? Select an answer ✓ (Please enter a decimal) (please show your answer to 3 decimal places.) c. The test statistic?v= d. The p-value = e. The p-value is ? ✓ a (Please show your answer to 4 decimal places.)
In a volunteer group, adults 21 and older volunteer between 1 and 9 hours each week. The program includes people from a community college, a four-year college, and non-students. The leader of the group is interested in seeing if there is a relationship between the type of volunteer and the number of hours volunteered. She collected the following data below. Perform a hypothesis test to help her at 0.01 level of significance. 1-3 hours 4-6 hours 7-9 hours Community College Student 97 100 64 Four-Year College Student 99 128 55 Non-Student 99 138 41 What are the expected numbers? (round to 3 decimal places) 1-3 hours 4-6 hours 7-9 hours Community College Student Four-Year College Student Non-Student The hypotheses are H0:H0: The number of hours volunteered is independent of the type of volunteer. HA:HA: The number of hours volunteered is dependent of the type of volunteer.(claim) Since αα = 0.01 the critical value is 13.277 The test…
(a) Determine the null and alternative hypotheses that would be used to test the effectiveness of the marketing campaign. Ho: = μ 56 H₁: μ > 56 (Type integers or decimals. Do not round.) (b) A sample of 884 Americans provides enough evidence to conclude that marketing campaign was effective. Provide a statement that should be put out by the marketing department. O A. There is not sufficient evidence to conclude that the mean consumption of popcorn has risen. B. There is sufficient evidence to conclude that the mean consumption of popcorn has risen. C. There is sufficient evidence to conclude that the mean consumption of popcorn has stayed the same. D. There is not sufficient evidence to conclude that the mean consumption of popcorn has stayed the same.
The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,941 kilowatt hours (kWh) of electricity this year. A local power company believes that residents in their area use more electricity on average than EIA's reported average. To test their claim, the company chooses a random sample of 115 of their customers and calculates that these customers used an average of 11,425 kWh of electricity last year. Assuming that the population standard deviation is 3217 kWh, is there sufficient evidence to support the power company's claim at the 0.02 level of significance? Step 3 of 3: Draw a conclusion and interpret the decision.
You are conducting a study to see if the proportion of voters who prefer the Democratic candidate is significantly larger than 74% at a level of significance of = 0.01. According to your sample, 51 out of 62 potential voters prefer the Democratic candidate. a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: Ho: ?> Select an answer (please enter a decimal) H₁ Select an answer (Please enter a decimal) c. The test statistic? = your answer to 3 decimal places.) d. The p-value = to 4 decimal places.) e. The p-value is ? ✰ a (please show (Please show your answer f. Based on this, we should Select an answer the null hypothesis. g. Thus, the final conclusion is that ... The data suggest the population proportion is not significantly larger than 74% at a = 0.01, so there is not sufficient evidence to conclude that the proportion of voters who prefer the Democratic candidate is larger than 74%. The data suggest the population proportion is not…
Suppose the national average dollar amount for an automobile insurance claim is $777.86. You work for an agency in Michigan and you are interested in whether or not the state average is less than the national average. The hypotheses for this scenario are as follows: Null Hypothesis: μ ≥ 777.86, Alternative Hypothesis: μ < 777.86. You take a random sample of claims and calculate a p-value of 0.0232 based on the data, what is the appropriate conclusion? Conclude at the 5% level of significance. Question 9 options: 1) The true average claim amount is significantly different from $777.86. 2) The true average claim amount is significantly less than $777.86. 3) The true average claim amount is significantly higher than $777.86. 4) We did not find enough evidence to say the true average claim amount is less than $777.86. 5) The true average claim…
If, in a one-tail hypothesis test where H0 is only rejected in the upper tail, the p-value=0.6792 and ZSTAT=−0.47,what is the statistical decision if the null hypothesis is tested at the 0.07 level of significance? What is the statistical decision? Since the p-value is Less than or Greater than or equal to a= , Reject , Do not reject H0.
An elementary school claims the variation in reading scores of its fourth grade students is less than o = 3. 25. Determine whether the hypothesis for this claim is left-tailed, right-tailed, or two-tailed. left-tailed right-tailed two-tailed
Assume that a simple random sample has been selected from a normally distributed population and test the given claim.A researcher wants to test the claim that convicted burglars spend an average of 18.7 months in jail. She takes a random sample of 11 such cases from court files and finds that x = 21.3 months and s = 5.7 months. Test the claim at the 0.05 significance level.State the claim in words: Convicted Burglars spend an average of 18.7 months in jail. Ho μ = 18.7 H1 μ ≠ 18.7 What calculator function is needed? p value = __________ Since p ________ α, ________________________ the null hypothesis. Final conclusion:

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