Question

A sample of 52 Elementary Statistics students includes 13 women. Assuming the sample is 4. random. . . (a) Estimate the percentage of women taking Elementary Statistics with 98% confidence. (b) At 10% significance, test whether less than 40% of the enrollment in all Elementary Statistics classes consists of women. (c) If in fact 46% of the students in Elementary Statistics classes are women, find the power of the above test in detecting this parameter.

Answer 1

(a) The 98% Confidence Interval for the proportion:

= 13/52 = 0.25, 1 - = 0.75, n = 52, = 0.02

The Zcritical (2 tail) for = 0.02, is 2.326

The Confidence Interval is given by ME, where

The Lower Limit = 0.25 - 0.14 = 0.11

The Upper Limit = 0.25 + 0.14 = 0.39

The Confidence Interval is (0.11 , 0.39)

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(b) Hypothesis Test for a single proportion:

p = 0.4, = 0.25, n = 52, = 0.10

The Hypothesis:

H0: p = 0.4

Ha: p < 0.4

This is a Left tailed Test.

The Test Statistic:

The p Value: The p value (Left tail) for Z = -2.21, is; p value = 0.0136

The Critical Value: The critical value (Left tail) at = 0.10, Zcritical = -1.282

The Decision Rule:

The Critical Value Method: If Zobserved is <- Zcritical Then Reject H0.

The p-value Method: Also If the P value is < , Then Reject H0

The Decision:   

The Critical Value Method: Since Z observed (-2.21) is < -Zcritical (-1.282), We Reject H0.

The p-value Method: Since P value (0.0136) is < (0.10), We Reject H0.

The Conclusion: There is sufficient evidence at the 90% significance level to conclude that the proportion of women enrolling in all elementary statistics classes is less than 40%.

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(c) Power of the Test:

Hypothesized proportion = 0.4

True Proportion = 0.46

The test statistic is given by:

Step 1: The value of for which the null hypothesis would get rejected, with the hypothesized proportion of 0.4

would get rejected if Z - 1.282

= p + Z * SQRT [p * (1-p)/n]

= 0.4 + (-1.282) * SQRT(0.4 * 0.6/52) = 0.4 - 0.087 = 0.313

Step 2: To find the probability at which = 0.313 gets rejected, when the true population mean = 0.46

Z = (0.313 - 0.46) / [SQRT(0.46 * 0.54/52)] = 2.128

The probability, P(X < 0.313) at Z = 2.128 is the power of the test = 0.0167

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