Question

In: Statistics and Probability

In a random sample of 340 cars driven at low altitudes, 46 of them exceeded a...

In a random sample of 340 cars driven at low altitudes, 46 of them exceeded a standard of 10 grams of particulate pollution per gallon of fuel consumed. In an independent random sample of 85 cars driven at high altitudes, 21 of them exceeded the standard. Can you conclude that the proportion of high altitude vehicles exceeding the standard is greater than the proportion of low altitude vehicles exceeding the standard?

a. State whether the test is:

i) a two-sample t-test (independent samples)

ii) a matched pairs

iii) a two sample proportion test

b. Write H0 and H1

c. Using Minitab, list the test statistic the p-value your conclusion: reject H0 or do not reject H0. Note: if α is not provided, use a 0.05 significance level

d.   Write a sentence that explains your conclusion in context with the claim. Include the significance level and p-value in this sentence.

e.   Copy and paste the relevant Minitab output into the document. Answers alone are sufficient, you do not need to copy the exercise into the document.

Solutions

Expert Solution

a) iii) a two sample proportion test

_________________________________

    

High altitude (1) Low altitude (2)

_______________________________

c) Test - stats (z) = 2.22

   P-value = 0.0132

    P-value is less than 0.05

   So, Reject Ho

________________________________

d) Since, P-value is less than the significance level so, reject the null hypothesis

we have enough evidence to support the claim that the high altitude proportion is greater than the low altitude proportion

________________________________

e)

Test and CI for Two Proportions

Method

p₁: proportion where Sample 1 = Event
p₂: proportion where Sample 2 = Event
Difference: p₁ - p₂

Descriptive Statistics

Sample N Event Sample p
Sample 1 85 21 0.247059
Sample 2 340 46 0.135294

Estimation for Difference

Difference 95% Lower
Bound for
Difference
0.111765 0.028988

CI based on normal approximation

Test

Null hypothesis H₀: p₁ - p₂ = 0
Alternative hypothesis H₁: p₁ - p₂ > 0
Method Z-Value P-Value
Normal approximation 2.22 0.0132

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