An educator claims that the dropout rate for freshmen in a certain college is mo

Question

An educator claims that the dropout rate for freshmen in a certain college is more than 20%. Last year, from a random sample of 200 freshmen students 49 withdrew.

A. The null hypothesis would not be rejected at any level of significance.

B. At a level of significance of 0.05 the null hypothesis would be rejected, but not at 0.1

C. The null hypothesis would not be rejected at either level of significance, 0.1 and 0.05

D. At a level of significance of 0.1 the null hypothesis would be rejected, but not at 0.05

Explanation / Answer

this is a case of population proportion testing .

given that P = 0.2 and p = 49/200 = 0.245

The test statistic is a z-score (z) defined by the following equation.

z = (p - P) /

Compute the standard deviation () of the sampling distribution.

= sqrt[ P * ( 1 - P ) / n ]

so sqrt(0.2*(1-0.2)/200)= 0.0282

putting the value in the z formula = (0.245-0.2)/0.0282 = 1.595

now lets use the z table to check the z critical values at 0.05 and 0.1

for 0.05 = -1.6449

for 0.1 = -1.2816

as in both the cases , the calculated Z score is greater than the Z critical value , hence we reject null in favor of altenate hypothesis.

hence it would be rejected in both the cases

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