Question

You are conducting a study to see if the probability of catching the flu this year is significantly more than 0.13. You use a significance level of α=0.10α=0.10.

      H0:p=0.13H0:p=0.13
      H1:p>0.13H1:p>0.13

You obtain a sample of size n=692n=692 in which there are 103 successes.

What is the test statistic for this sample?
test statistic = (Report answer accurate to 3 decimal places.)

What is the p-value for this sample?
p-value = (Report answer accurate to 4 decimal places.)

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 more than 0.13.
• There is not sufficient evidence to warrant rejection of the claim that the probability of catching the flu this year is more than 0.13.
• The sample data support the claim that the probability of catching the flu this year is more than 0.13.
• There is not sufficient sample evidence to support the claim that the probability of catching the flu this year is more than 0.13.

Answer 1

Solution :

Given that ,

n = 692

x = 103

The null and alternative hypothesis is

H0 : p = 0.13

Ha : p > 0.13

This is the right tailed test .

= x / n = 103 / 692 = 0.1488

P0 = 0.13

1 - P0 = 1 - 0.13 = 0.87

Test statistic = z

= - P0 / [P0 * (1 - P0 ) / n]

= 0.1488 - 0.13 / [0.13 ( 1 - 0.13 ) / 692 ]

= 1.474

The test statistic = 1.474

P-value = 0.0702

= 0.10

0.0702   0.10  

P-value     

P - value is less than ( or equal to )

Reject the null hypothesis .

Conclusion : - There is sufficient evidence to warrant rejection of the claim that the probability of catching the flu this year is more than 0.13.