Question

> Toastmasters International cites a report by Gallup Poll that 40% of Americans fear public speaking. A student believes that less than 40% of students at her school fear public speaking. She randomly surveys 361 schoolmates and finds that 144 report they fear public speaking. Conduct a hypothesis test at the 5% level to determine if the percent at her school is less than 40%.

> Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)
> A. State the distribution to use for the test. (Round your answers to four decimal places.)
> B. What is the test statistic? (If using the z distribution round your answers to two decimal places, and if using the t distrib
> C. What is the p-value? (Round your answer to four decimal places.)
> D. Construct a 95% confidence interval for the true proportion. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your answers to four decimal places.)

Answer 1

Solution:

p = proportion of Americans fear public speaking = 0.40

n = sample size = 361

x = Number of schoolmates  fear public speaking = 144

Level of significance = 0.05

We have to test the hypothesis that  less than 40% of students at her school fear public speaking.

that is: p < 0.40

Part A. State the distribution to use for the test.

For testing hypothesis for proportion , we use normal (z) distribution.

thus

Part B. What is the test statistic?

where

thus

Part C) What is the p-value?

For left tailed test , p-value is:

p-value = P(Z < z test statistic)

p-value = P(Z < -0.04)

Look in z table for z = -0.0 and 0.04 and find corresponding area.

P( Z< -0.04) =0.4840

thus

p-value = P(Z < -0.04)

p-value = 0.4840

Part D. Construct a 95% confidence interval for the true proportion. Label the point estimate and the lower and upper bounds of the confidence interval.

The point estimate is:

and

where

We need to find z_c value for c=95% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.95) /2 = 1.95 / 2 = 0.9750

Look in z table for Area = 0.9750 or its closest area and find z value.

Area = 0.9750 corresponds to 1.9 and 0.06 , thus z critical value = 1.96

That is : Z_c = 1.96

Thus

Thus

and