Question

Suppose that a sample of n=400 students at the same university​ (instead of n=200​) determines that...

Suppose that a sample of n=400 students at the same university​ (instead of n=200​) determines that 25​% of the sample use the web browser. At the .05 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 ​20.5%?

a. Calculate the test statistic for the second sample.

b. What is the​ p-value?

c. State the conclusion of the test. _____ the null hypothesis. There is _____ evidence to conclude that the market share at the university is ____ the worldwide market share of ​20.5%.

Homework Answers

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
The worldwide market share for a web browser was 20.5​% in a recent month. Suppose that...
The worldwide market share for a web browser was 20.5​% in a recent month. Suppose that a sample of 200200 random students at a certain university finds that 5050 use the browser. Suppose that a sample of n=800 students at the same university​ (instead of n=200​) determines that 25​% of the sample use the web browser. At the 0.05 level of​ significance, is there evidence that the market share for the web browser at the university is greater than the...
The worldwide market share for a web browser was 20.5​% in a recent month. Suppose that...
The worldwide market share for a web browser was 20.5​% in a recent month. Suppose that a sample of 200 random students at a certain university finds that 50 use the browser. Complete parts​ (a) through​ (d) below. a. At the 0.05 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 20.5​%? Determine the null and alternative hypotheses. A. H0​: piequals0.205​; H1​: pinot equals0.205...
From a random sample from normal population, we observed sample mean=84.5 and sample standard deviation=11.2, n...
From a random sample from normal population, we observed sample mean=84.5 and sample standard deviation=11.2, n = 16, H0: μ = 80, Ha: μ < 80. State your conclusion about H0 at significance level 0.01. Question 2 options: Test statistic: t = 1.61. P-value = 0.9356. Reject the null hypothesis. There is sufficient evidence to conclude that the mean is less than 80. The evidence against the null hypothesis is very strong. Test statistic: t = 1.61. P-value = 0.0644....
An article in the San Jose Mercury News stated that students in the California state university...
An article in the San Jose Mercury News stated that students in the California state university system take 5.5 years, on average, to finish their undergraduate degrees. A freshman student believes that the mean time is less and conducts a survey of 60 students. The student obtains a sample mean of 4.9 with a sample standard deviation of 0.3. Is there sufficient evidence to support the student's claim at an α=0.1 significance level? Preliminary: Is it safe to assume that...
The mean GPA at a certain large university is 2.80. A random sample of 20 business...
The mean GPA at a certain large university is 2.80. A random sample of 20 business students found a mean GPA of 2.95 with a standard deviation of 0.31. Is there sufficient evidence to show that the mean GPA for business students is greater than the mean GPA for the university? a) state the hypothesis b) calculate the test statistic c) find the p-value d) make a statistical decision at a level of significance of 0.10 e) state your conclusion...
An article in the San Jose Mercury News stated that students in the California state university...
An article in the San Jose Mercury News stated that students in the California state university system take 6 years, on average, to finish their undergraduate degrees. A freshman student believes that the mean time is less and conducts a survey of 39 students. The student obtains a sample mean of 5.7 with a sample standard deviation of 1.8. Is there sufficient evidence to support the student's claim at an α=0.01α=0.01 significance level? Preliminary: Is it safe to assume that...
A local university requires that all first year students complete a math course during first semester....
A local university requires that all first year students complete a math course during first semester. This year the university is evaluating a new online version of the course. A random sample of n = 20 students is selected and the students are placed in the online course. At the end of the semester, all students take the same math exam. The average score for the sample of n = 20 students is M = 85. For the general population...
70% of students at a university live on campus. A random sample found that 31 of...
70% of students at a university live on campus. A random sample found that 31 of 50 male students and 41 of 55 of female students lived on campus. At the .05 level of significance, is there sufficient evidence to conclude that a difference exists between the proportion of male students who live on campus and the proportion of female students who live on campus?
Mean 3407.94 Median 2364.00 Standard Deviation 3988.42 Minimum 136.00 Maximum 21451.00 n=52 Determine if there is...
Mean 3407.94 Median 2364.00 Standard Deviation 3988.42 Minimum 136.00 Maximum 21451.00 n=52 Determine if there is sufficient evidence to conclude the average amount of marriages is greater or equal to 7000 in the United States and territories at the .05 level of significance. Clearly state a null and alternative hypothesis Give the value of the test statistic Report the P-Value Clearly state your conclusion (Reject the Null or Fail to Reject the Null) Explain what your conclusion means in context...
summary for marriages Mean 3407.94 Median 2364.00 Standard Deviation 3988.42 Minimum 136.00 Maximum 21451.00 n=52 Determine...
summary for marriages Mean 3407.94 Median 2364.00 Standard Deviation 3988.42 Minimum 136.00 Maximum 21451.00 n=52 Determine if there is sufficient evidence to conclude the average amount of marriages is greater or equal to 9000 in the United States and territories at the .05 level of significance Clearly state a null and alternative hypothesis Give the value of the test statistic Report the P-Value Clearly state your conclusion (Reject the Null or Fail to Reject the Null) Explain what your conclusion...