You are testing the null hypothesis that mu = 0 versus the alternative mu > 0 us

Question

You are testing the null hypothesis that mu = 0 versus the alternative mu > 0 using alpha = .05. Assume sigma = 18. Suppose x = 5.3 and n = 15. Calculate the test statistic and its P-value. Repeat assuming the same value of x but with n = 25. Do the same for sample sizes of 35, 45, and 55. (Round the test statistic to two decimal places. Round the P-value to four decimal places.) n = 15: z P- value n = 25: z P- value n = 35: z P- value n = 45: z P- value n = 55: z P-value Plot the values of the test statistics versus the sample size. Do the same for the P-values. (Do this on paper. Your instructor may ask you to turn this in.) Summarize what this demonstration shows about the effect of the sample size on significance testing. As sample size increases, a test becomes more significant. As sample size increases, a test becomes less significant. As sample size increases, there is no effect on the significance. As sample size decreases, a test becomes more significant.

Explanation / Answer

H0 : = 0

Ha : > 0

xbar = 5.3 and = 18 and n = 15

so standard error of mean = /n = 18/ 15 = 4.6475

so Test statistic

Z = (xbar -  H)/(/n) = 5.3/ 4.6475 = 1.14

so critical value of Z = 1.645

P - value = Pr(xbar > 5.3 ; 0; 4.6475) =  (1.14)

where   is the normal cumulative distribution function

so p- value = 0.1271 >0.05

so here Z < Zcritical so we cannot reject the null hypothesis and say population mean = 0

(b) n = 25

so standard error of mean = /n = 18/ 25 = 3.6

so Test statistic

Z = (xbar -  H)/(/n) = 5.3/ 3.6 = 1.47

so critical value of Z = 1.645 for alpha = 0.05

P - value = Pr(xbar > 5.3 ; 0; 3.6) =  (1.47)

where   is the normal cumulative distribution function

so p- value = 0.07 >0.05

so here Z < Zcritical so we cannot reject the null hypothesis and say population mean = 0

(c)

n = 35

so standard error of mean = /n = 18/ 35 = 3.042

so Test statistic

Z = (xbar -  H)/(/n) = 5.3/ 3.042 = 1.74

so critical value of Z = 1.645 for alpha = 0.05

P - value = Pr(xbar > 5.3 ; 0; 3.6) =  (1.74)

where   is the normal cumulative distribution function

so p- value = 0.041 < 0.05

so here Z < Zcritical so we canreject the null hypothesis and say population mean > 0

(d)

n = 45

so standard error of mean = /n = 18/ 45 = 2.683

so Test statistic

Z = (xbar -  H)/(/n) = 5.3/ 2.683 = 1.975

so critical value of Z = 1.645 for alpha = 0.05

P - value = Pr(xbar > 5.3 ; 0; 3.6) =  (1.74)

where   is the normal cumulative distribution function

so p- value = 0.024 < 0.05

so here Z < Zcritical so we canreject the null hypothesis and say population mean > 0

(d)

n = 55

so standard error of mean = /n = 18/ 55 = 2.427

so Test statistic

Z = (xbar -  H)/(/n) = 5.3/ 2.427 = 2.184

so critical value of Z = 1.645 for alpha = 0.05

P - value = Pr(xbar > 5.3 ; 0; 3.6) =  (1.74)

where   is the normal cumulative distribution function

so p- value = 0.0145 < 0.05

so here Z < Zcritical so we canreject the null hypothesis and say population mean > 0

As sample size increases, a test becomes more significant. (option A)

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