Question 2-Hypothesis test (8 marks) Research Question: Is the average usage of
ID: 3049674 • Letter: Q
Question
Question 2-Hypothesis test (8 marks) Research Question: Is the average usage of smartphones by MUIC students equal to 95 minutes per day? MUIC Marketing Services in 2016, randomly selected 50 students. Suppose the population standard deviation is unknown, we can use the sample standard deviation and results of the study which are shown below: Histogram for smartphone user Key descriptive statisties Mean Standard Deviation Count 98.0968 19.4769 50 4 600 100 120 140 Use the statistical summaries shown above to determine whether the research claim is true. Hypothesis test: (8 marks: each question worth 1 mark) H | Null and Alternative Hypothesis : Ho: = A Assumption: From the , the normality assumption appears reasonable. The sample mean can be assumed to be taken from a Normal distribution. 9- s/Vn with df L Test statistic (4dpit p p-value (Circle the correct option): a. TDIST.2T(1.1243, 50)=0.2663 c. T.DIST.2T (1.1243, 49)-0.2664 b. T.DIST 2T (1.1130, 50)-0.2710 d. T.DIST 2T (1.1130, 49)-0.2711 Decision (Circle the correct option): Since p-value> 0.05, do not reject Ho Since p-valueExplanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: = 95
Alternative hypothesis: 95
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n)
S.E = 2.754
DF = n - 1 = 50 - 1
D.F = 49
t = (x - ) / SE
t = 1.12
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
Since we have a two-tailed test, the P-value is the probability that the t statistic having 49 degrees of freedom is less than -1.12 or greater than 1.12.
(c) T.DIST.2T(1.1243,49) = 0.2664
Thus, the P-value = 0.2664
(a) Since p-value > 0.05, do not rejec H0.
Interpret results. Since the P-value (0.2682) is greater than the significance level (0.05), we have to accept the null hypothesis.
(c) The average smartphone use could be 95 minutes per day, as claimed by the researcher.
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