Question

In a recent survey, 56​% of employed adults reported that basic mathematical skills were critical. A supervisor thinks this percentage has increased due to increased use of technology in the workplace. She takes a random sample of 470 employed adults and finds that 281 of them feel that basic mathematical skills are critical or very important to their job. Test her hypothesis at the α=0.05 level of significance.

-What is the test statistics?

-What is the critical value?

-Estimate the left boundary of a 95% confidence interval of the population proportion

Answer 1

Solution :

Given that,

= 0.56

1 - = 0.44

n = 470

x = 281

Level of significance = = 0.05

Point estimate = sample proportion = = x / n = 0.60

This a right (One) tailed test.

The null and alternative hypothesis is,

Ho: p = 0.56

Ha: p 0.56

Test statistics

z = ( - ) / *(1-) / n

= ( 0.60 - 0.56) / (0.56*0.44) / 470

= 1.65

Critical value of  the significance level is α = 0.05, and the critical value for right-tailed test is

= 1.64

Since it is observed that , z = 1.65 > = 1.64 , it is then concluded that the null hypothesis is rejected.

P-value = P(Z>z)

= 1 - P(Z <z )

= 1- P(Z < 1.65)

= 1 - 0.9505

= 0.0495

The p-value is p = 0.0495, and since p = 0.0495 < 0.05, it is concluded that the null hypothesis is rejected.

Conclusion:

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that A supervisor thinks this percentage has increased due to increased use of technology in the workplace at the α = 0.05 significance level.

The 95% confidence interval for p is: 0.554 < p < 0.642.