1. Proportion of Female Students in STT-215 According to US News and World Repor

Question

1. Proportion of Female Students in STT-215

According to US News and World Report, the proportion of females attending UNCW is 0.62 (62%). At the 0.05 level of significance, is there evidence that the proportion of females taking STT-215 (assume a random sample) is greater than the proportion of females attending UNCW?

Report p-value and use it to support your conclusion

2. Proportion of Female Students in STT-215 and Bentley University

Compare the proportion of females from samples of Bentley University and UNCW students. Is the proportion of females at UNCW significantly greater than the proportion of females at Bentley? Use a 0.05 level of significance.

Report p-value and use it to support your conclusion

BU Gender UNCW Gender M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M F M F M F M F M F M F M F M F M F M F M F M F M F M F M F M F M F M F M F F F F F F F F F F F F F F F F F F F F F F F M F M F M F M F M F M F M F M F M F M F M F M F

Explanation / Answer

Question -1

This question is Hypothesis testing for one sample proportion.

H0 : p = 0.62

HA  : p > 0.62

The test is one - tailed test

sample size = 62

Here sample prorpotion p^ from the given sample is:

p^ = 42/62 = 0.6774

here, standard error of proportion = sqrt [p^ * (1-p^)/N] = sqrt [0.6774 * 0.3226/62] = 0.0594

Test statistic

Z = (p^ - 0.62)/ 0.0594 = (0.6774 -0.62)/ 0.0594 = 0.9663

so p - value= Pr(Z > 0.9663) = 1- 0.833 = 0.167

so p - value is higher then the standard significant level 0.05 so we shall no reject the nul hypothesis and can conclude that  proportion of females taking STT-215 is not greater than the proportion of females attending UNCW.

Question 2

H0 : pBU = pUNCW

Ha : pBU < pUNCW

Here p^ (BU) = 11/62 = 0.1774

so Here Test statistic

pooled estimate p = (11 + 42)/ (62 + 62) = 0.4274

se (p1 - p2) = sqrt [ p * (1-p) * (1/n1 + 1/n2 )] = sqrt [0.4274 * 0.5726 * (1/62 + 1/62) ]

= 0.08885

Z - statistic

Z = (p1 - p2)/ se(p1 - p2) = (0.6774 - 0.1774)/ 0.08885 = 5.6274

Here

p - value= Pr( Z > 5.6274) = 0.00000

so here we shall reject the null hypothesis at alpha = 0.05 leve and can conclude that the proportion .of females at UNCW significantly greater than the proportion of females at Bentley

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