Question

You are conducting a study to see if the probability of catching the flu this year is significantly different from 0.77. You use a significance level of α=0.005α=0.005.

      H0:p=0.77H0:p=0.77
      H1:p≠0.77H1:p≠0.77

You obtain a sample of size n=489n=489 in which there are 386 successes.

What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic =

What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value =

The p-value is...

• less than (or equal to) αα
• greater than αα

This test statistic leads to a decision to...

• reject the null
• accept the null
• fail to reject the null

As such, the final conclusion is that...

• There is sufficient evidence to warrant rejection of the claim that the probability of catching the flu this year is different from 0.77.
• There is not sufficient evidence to warrant rejection of the claim that the probability of catching the flu this year is different from 0.77.
• The sample data support the claim that the probability of catching the flu this year is different from 0.77.
• There is not sufficient sample evidence to support the claim that the probability of catching the flu this year is different from 0.77.

Answer 1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H₀ : p = 0.77

H_a : p 0.77

= x / n = 386 / 489= 0.7894

P₀ = 0.7

1 - P₀ = 0.23

Test statistic = z

= - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.7894 - 0.77 / [(0.77 * 0.23) / 489]

= 1.018

P(z > 1.018) = 1 - P(z < 1.018) = 0.1543

P-value = 2 * 0.1543 = 0.3086

= 0.05

P-value >

fail to reject the null

There is not sufficient evidence to warrant rejection of the claim that the probability of catching the flu this year is different from 0.77.