Question

1. A researcher is testing the claim that adults consume an average of at least 1.85 cups of coffee per day. A sample of 35 adults shows a sample mean of 1.70 cups per day with a sample standard deviation of 0.4 cups per day. Test the claim at a 5% level of significance. What is your conclusion?

2. A government Bureau claims that more than 50% of U.S. tax returns were filed electronically last year. A random sample of 150 tax returns for last year contained 86 that were filed electronically. Test the Bureau's claim at a 5% level of significance. What is your conclusion? Report the p-value for this test.

3. A major automobile company claims that its New electric powered car has an average range of more that 100 miles. A random sample of 50 new electric cars was selected to test the claim. Assume that the population standard deviation is 12 miles. A 5% level of significance will be used for the test.

    A) What would be the consequences of making a Type II error in this problem?

    B) Compute the Probability of making a Type II error if the true population mean is 105 miles.

    C) What is the maximum probability of making a Type I error in this problem?

Please Note: A hypothesis test answer must contain: a Null and an Alternate Hypothesis, a computed value of the test statistic, a critical value of the test statistic, a Decision , and a Conclusion.

Answer 1

1)

Given,

For Sample

Mean (x) = 1.7

Std Dev (s) = 0.4

n = 35

alpha = 0.05

Null and Alternate Hypothesis

H₀: µ = 1.85

H_a: µ > 1.85

Test Statistic

t = (x- µ₀)/ (s/n^1/2) = (1.7-1.85) / (0.4/35^1/2) = -2.22

p-value = TDIST(2.22,35-1,1) = 0.017

Result

Since the p-value is less than 0.05, the data is statistically significant and we reject the null hypothesis in favour of alternate hypothesis.

Conclusion

Adults consume an average of at least 1.85 cups of coffee per day

2)

Given,

For Sample

n = 150

Proportion (p) = 86/150 = 0.573

Std Dev (s) = {p*(1-p)/n}^1/2 = 0.04

alpha = 0.05

Null and Alternate Hypothesis

H₀: p = 0.5

H_a: p > 0.5

Test Statistic

t = (p- p₀)/s = (0.573-0.5) / 0.04 = 1.82

p-value = TDIST(1.82,150-1,1) = 0.036

Result

Since the p-value is less than 0.05, the data is statistically significant and we reject the null hypothesis in favour of alternate hypothesis.

Conclusion

More than 50% of U.S. tax returns were filed electronically last year

3)

a)

A Type II Error is the non-rejection of a false null hypothesis ie We concluded that company’s New electric powered car’s average is greater than 100 mph when in fact it is not greater than 100 mph