Question

The school paper claims that 35% of students have credit card debt. But one thinks it’s more. simple random sample “196” students and “78,” say that they have credit card debt.

Is there sufficient evidence at the 5% significance level (α = .05) to indicate that the percent of students with credit card debt is more than 35%

Show all work (or explain what is plug into the calculator). Use specific wording.

Ho:

Ha:

Test statistic:

p-value:

Statistical conclusion:

Conclusion in plain English:

Answer 1

Solution :

Given that,

= 0.35

1 - = 0.65

n = 196

x = 78

Level of significance = = 0.05

Point estimate = sample proportion = = x / n = 0.398

This a two- tailed test.

This a right (One) tailed test.

The null and alternative hypothesis is,

Ho: p = 0.35

Ha: p 0.35

Test statistics

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

= ( 0.398 - 0.35) / (0.35*0.65) / 196

= 1.409

P-value = P(Z > z )

= 1 - P(Z < 1.409 )

= 1 - 0.9206

= 0.0794

The p-value is p = 0.0794, and since p = 0.0794 > 0.05, it is concluded that fail to reject the null hypothesis.

There is not sufficient evidence at the 5% significance level (α = .05) to indicate that the percent of students with credit card debt is more than 35%