There are 1440 minutes in a day. Do Cornell students spend more of their preciou
ID: 3340953 • Letter: T
Question
There are 1440 minutes in a day. Do Cornell students spend more of their precious free minutes on social media, or on face-to-face human socialization? Assume that the number of minutes a student spends on social m edia in one day is from a normal population with m ean 1-The number of minutes a student spends on in-person socialization is norm ally distributed with mean 2 We set up an experiment in which, during one week day, we record the number of minutes spent on either of the "social" activities for a group of 10 students. Suppose we find a sample mean of 116 minutes spent on social media, and 34 minutes spent on in-person socialization. In particular, we have the following pairwise data: Student 1 2|3 4 567 89 10 l media 0 20 46 95 112 116 142 144 160320 In-person 62 25 30 34 36127 31575 (a) Construct a 99% confidence interval for 12 (b) Consider a hypothesis test of Explain in words what the null and alternative hypotheses mean in this setting. for the hypothesis test with paired data, the test statistic is (c) Carry out the hypothesis test described in part (b) with a significance level of -0.01. In general, where D and sp are the sample mean and sample standard deviation of the differences in the prelim scores. The rejection region for such a test is t2 ta,n-1 where t is the observed value of the test statistic T and ton-1 is the critical value from a t distribution with n-1 degrees of freedom. What do you conclude from the result of the hypothesis test? (d) What are the meanings of Type I and Type II error in this setting?Explanation / Answer
1) The confidence interval for 1 - 2 is evaluated as follows:
1 - 2 = (M1 - M2) ± ts(M1 - M2)
where M1 is meaan of sample1
M2 is meaan of sample2
t is t statistic determined by confidence level
s(M1 - M2) = standard error = ((s2p/n1) + (s2p/n2))
where sp is the pooled variance
Pooled variance
s2p = (SS1 + SS2) / (df1 + df2) = 8541 / 18 = 474.5
Standard Error
s(M1 - M2) = ((s2p/n1) + (s2p/n2)) = ((474.5/10) + (474.5/10)) = 9.74
1 - 2 = (M1 - M2) = 82
Confidence Interval
1 - 2 = (M1 - M2) ± ts(M1 - M2) = 82 ± (2.88 * 9.74) = 82 ± 28.04
Thus the 99% CI is [53.96, 110.04] i.e we are 99% confident that the difference between your two population means (1 - 2) lies between 53.96 and 110.04.
2) Here, the null hypothesis 1 - 2 = 0 implies that there is no difference between the means of two populations in the sample
and the alternate hypothesis 1 - 2 > 0 implies that the sample mean of first population is greater than the sample mean of second population
3) Since the t value is greater than ta,n-1
=> The test is significant and the null hypothesis cannot be rejected
4) Type 1 error is the probability of rejecting the null hypothesis when the null hypotheis is true.
Type 2 error is the probability of accepting the null hypothesis when the null hypotheis is false.
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