Question

A random sample of 40 binomial trials resulted in 16 successes. Test the claim that the population proportion of successes does not equal 0.50. Use a level of significance of 0.05.

(a) Can a normal distribution be used for the p̂ distribution? Explain.

Yes, np and nq are both less than 5.No, np is greater than 5, but nq is less than 5.     No, nq is greater than 5, but np is less than 5.Yes, np and nq are both greater than 5.No, np and nq are both less than 5.

(b) State the hypotheses.

H₀: p = 0.5; H₁: p > 0.5H₀: p = 0.5; H₁: p ≠ 0.5     H₀: p < 0.5; H₁: p = 0.5H₀: p = 0.5; H₁: p < 0.5

(c) Compute p̂.


Compute the corresponding standardized sample test statistic. (Round your answer to two decimal places.)


(d) Find the P-value of the test statistic. (Round your answer to four decimal places.)


(e) Do you reject or fail to reject

H₀?

Explain.

At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.     At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.

(f) What do the results tell you?

The sample p̂ value based on 40 trials is not sufficiently different from 0.50 to not reject H₀ for α = 0.05.The sample p̂ value based on 40 trials is sufficiently different from 0.50 to not reject H₀ for α = 0.05.     The sample p̂ value based on 40 trials is sufficiently different from 0.50 to justify rejecting H₀ for α = 0.05.The sample p̂ value based on 40 trials is not sufficiently different from 0.50 to justify rejecting H₀ for α = 0.05.

Answer 1

Solution :

a) Yes, np and nq are both greater than 5

This is the two tailed test .

b) The null and alternative hypothesis is

H0 : p = 0.50

Ha : p 0.50

c) = x / n = 16 /40 = 0.40

P0 = 0.50

1 - P0 = 1 -0.50 = 0.50

Test statistic = z

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

= 0.40 - .50/ [0.50 *(0.50) /40 ]

= -1.26

d) P(z <-1.26 ) = 0.2076

P-value = 0.2076

= 0.05

0.2076 > 0.05

e) At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.

f) The sample p̂ value based on 40 trials is not sufficiently different from 0.50 to not reject H0 for α = 0.05.