Five years ago a study concerning the types of passenger vehicles driven on U.S.

Question

Five years ago a study concerning the types of passenger vehicles driven on U.S. roads found that 40% of those vehicles were SUVs. You suspect that in this area that that value has been exceeded. You randomly tally vehicles on various interstates in this area at various times and find that of the 84 vehicles you tally, 41 of them are SUVs. At the 0.054 level of significance, is your suspicion supported by the collected data?

Hypothesis Test Requirement

Response for this Hypothesis Test

Null Hypothesis:

Alternative Hypothesis:

Critical Value(s):

Critical Region:

Test Value:

p-value:

Statistical Decision:

Justification for Decision:

Decision in Contextual Terms:

Justification for Distribution Used:

Hypothesis Test Requirement

Response for this Hypothesis Test

Null Hypothesis:

Alternative Hypothesis:

Critical Value(s):

Critical Region:

Test Value:

p-value:

Statistical Decision:

Justification for Decision:

Decision in Contextual Terms:

Justification for Distribution Used:

Explanation / Answer

Here, we have to use the z test for the population proportion:

The null and alternative hypothesis for this test is given as below:

Null hypothesis: H0: p = 0.40

Alternative hypothesis: Ha: p > 0.40

This is a one tailed test.

We are given a level of significance or alpha value = 0.054

Critical value = 1.6072

Critical Region = Z > 1.6072

Test statistic = (P – p) / sqrt(pq/n)

Where, P = X/n = 41/84 = 0.488095238

We have p = 0.40, q = 1 – p = 1 – 0.40 = 0.60 and n = 84

Test statistic = Z = (0.488095238 – 0.40) / sqrt(0.40*0.60/84)

Test Value = Z = 1.6481

P-value = 0.0497

Alpha value = 0.054

Statistical Decision:

So, we reject the null hypothesis

Justification for decision:

P-value < Alpha value

Test statistic value > Critical value

Decision in contextual terms:

We concluded that more than 40% of the vehicles are SUVs.

Justification for Distribution Used:

np = 84*0.40 = 33.6>5 and nq = 84*0.60 = 50.4>5, so we can use normal distribution or Z test statistic.

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