Question

A random sample of 60 binomial trials resulted in 18 successes. Test the claim that the population proportion of successes exceeds 18%. Use a level of significance of 0.01

Answer 1

We conclude that the population proportion of successes is less than or equal to 18%.

Explanation:

We are given that a random sample of 60 binomial trials resulted in 18 successes.

We have to test the claim that the population proportion of successes exceeds 18%.

Let p = population proportion of successes

So, Null Hypothesis,
`H_0` : p
`\leq` 18% {means that the population proportion of successes is less than or equal to 18%}

Alternate Hypothesis,
`H_A` : p > 18% {means that the population proportion of successes exceeds 18%}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. =
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` ~ N(0,1)

where,
`\hat p` = sample proportion of successes =
`(18)/(60)` = 0.30

n = sample of trials = 60

p = population proportion

So, test statistics =
`\frac{0.30-0.18}{\sqrt{(0.30(1-0.30))/(60) } }`

= 2.028

The value of the test statistics is 2.028.

Now at 0.01 significance level, the z table gives critical value of 2.3263 for right-tailed test. Since our test statistics is less than the critical values of z as 2.028 < 2.3263, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the population proportion of successes is less than or equal to 18%.