Question

10. A local college newsletter reported that the average American college student spends one hour (60 minutes) on a social media website daily. But you wonder if there is a differencs between males and femalas. Attached below is a sample of 60 users of the website (30 males and 30 females) and their recorded daily time spent on the website. Complete (a) and (b) below. Click the con to view the sample times. a Assuming that the variances n the population of times spent an the website per day are equal is there evidence a a difference n the mean time spent on the website per day between males and females? Use ล005 eve of significance Let μι be the mean daty tme spent on the website for male college students and μ2 be the mean daily t me spent on the website for female colege students. Determine thee hypotheses hoose the correct answer below. Determine the test statistic. -(Round to two decimal plaoes as needed.) Choose the correct concusion below. O A. Reject .There is sufficient evidence that the means differ. O B. Do not reject H. There is insufficient evidence that the means differ. ° C. Reject Ho . There is insufficent evidence that th归means differ O D. Do nat eject H.There is sufficient evidence that the means differ. b. In addition to equal variances, what other assumption is necessary in (a)? O A. In addition to equal variances. no other assumptions must be made because the sample sizes are large enough 30 for each sample). B. In addition to equal variances, it must be assumed that the samples are specifically chosen and nat O C. OD. independenty sampled. In addition to equal variances, it must be assumed that the population sizes are equal. In addition to equal variances, it must be assumed that the sample means are equal 1: Sample times Daily tme spent (in minutes) on the website for 30 males 210 930 45 30 10 5 720 330 60 150 150 150 210 210 180 45 120 270 30 150 20 45 45 150 60 15 45 30 150 100 Daily time spent (in minutes) on the website for 30 females 60 75 15 60 30 150 120 150 10 5 120 60 60 100 150 75 60 60 270 90 20 45 45 150 60 15 45 30 150 100

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u₁ = u ₂
Alternative hypothesis: u₁ $\neq$ u ₂

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 38.4501
DF = 58
t = [ (x₁ - x₂) - d ] / SE

t = 1.825

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 58 degrees of freedom is more extreme than -1.825; that is, less than -1.825 or greater than 1.825.

Thus, the P-value = 0.073

Interpret results. Since the P-value (0.073) is greater than the significance level (0.05), we have to accept the null hypothesis.

B) Do not reject H0, there is insufficient evidence that the mean differs.

b) A) In addidtion to equal variances, no other assumptions must be made because the sample sizes are large enough.