Question

In one satisfaction survey, 8023 subjects responses were collected. 24.94% of subjects reported being satisfied and the rest reported being unsatisfied. 1. Test the claim that the satisfied responses occur at a rate different than 25% using the 10% level of significance. 2. Find a 95% confidence interval for the population proportion of subjects being satisfied based on the survey data.

Answer #1

H0: p = 0.25

Ha: p 0.25

Sample proportion = 0.2494

Test statistics

z = - p / sqrt( P( 1 - P) / n)

= 0.2494 - 0.25 / sqrt( 0.25 * 0.75 / 8023)

= -0.12

This is test statistics value.

p-value = 2 * P( Z < -0.12)

= 2 * 0.4522

= 0.9044

At 0.10 level, since p-value > 0.10 level **we fail to
reject the null hypothesis**

b)

95% confidence interval for p is

- Z * sqrt( ( 1 - ) / n) < p < + Z * sqrt( ( 1 - ) / n)

0.2494 - 1.96 * sqrt( 0.2494 * 0.7506 / 8023) < P < 0.2494 + 1.96 * sqrt( 0.2494 * 0.7506 / 8023)

0.240 < p < 0.259

95% CI is **( 0.240 , 0.259)**

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