Question

Please write answer and formula ON PAPER and please don’t forget p value graph and calculator function and test statistic .

Stat 50 Final Exam Last Name First Name In a survey of 267 senior high school students, 43.1% said that, on average, they had less than 6 hours sleep each day. Test the claim that, on average, in the population of all senior students, the majority have less than 6 hours sleep each day. 9, Claim: Но: Ha: Random Variable and Distribution: Calculator function (including what you entered) and Test Statistic: P value and Graph: Decision: Conclusion:

Answer 1

Claim : In the population of all senior students, the majority have less than 6 hours sleep each day.

So null hypothesis is H₀ : P = 0.5

and alternative hypothesis is H_a : P > 0.5

Let X be the random variable which takes the value of number of students who less than 6 hours each day out of 267 students.

Here n = 267 and p = 0.50 ( under null hypothesis).

So that the sampling distribution of sample proportion (p) follows normal with mean P = 0.50 and standard deviation =

P(1- P) 0.50.5 0.0306 267

Formula of z test statistic:

p-value = P( Z > -2.2549) = 1 - P(Z < -2.2549) = "=1-NORMSDIST(-2.2549)" = 0.9879

Decision rule: (If level of significance is not given then take it as 0.05

1) If p-value < level of significance (alpha) then we reject null hypothesis

2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.

Here p value = 0.9879 > 0.05 so we used 2nd rule.

That is we fail to reject null hypothesis

Conclusion: At 5% level of significance there are not sufficient evidence to say that the sample data indicates in the population of all senior students, the majority have less than 6 hours sleep each day.

Let's use TI-84 calculator.

Command: STAT>>>TESTS>>>5:1-PropZTest...

P₀ : 0.5

X : 115 (X = 0.431*267 =115)

n: 267

Then select > P₀ ( because from the alternative hypothesis the test is right tailed test).

Then select Draw and click on Enter

So we get the following output: