Question

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Previously, an organization reported that teenagers spent 1.5 hours per day, on average, on the phone. However, now the organization thinks that, the mean is higher. 45 randomly chosen teenagers were asked how many hours per day they spend on the phone. The sample mean was 1.45 hours with a sample standard deviation of 0.30. Conduct a test using a significance level of α=0.05

a. The test statistic

b. The critical value

c. is there is enough evidence to say that teenager spend more time on the phone?

Answer 1

Given that previously, an organization reported that teenagers spent = 1.5 hours per day, on average, on the phone. However, now the organization thinks that the mean is higher. n = 45 randomly chosen teenagers were asked how many hours per day they spend on the phone. The sample mean was = 1.45 hours with a sample standard deviation of s= 0.30. To test the hypothesis at a significance level of α=0.05, we produce the hypotheses as:

Thus based on the hypothesis it will a left tailed test, but since the population standard is unknown hence t-distribution is applicable for hypothesis testing.

a) Test Statistic:

b) Critical value:

The critical value for the given significance level and type of hypothesis is calculated using the excel formula for t-distribution which takes the significance level and the degree of freedom n-1 as the parameters.

The formula used is =T.INV.2T(0.05, 44), thus tc computed as -1.68

So, reject the Ho if t <tc

c) Conclusion:

Since The test statistic is greater than tc hence the test statistic does not lie in the rejection region, hence we fail to reject the null hypothesis and conclude that there is insufficient evidence to support the claim.