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aleksandrvk [35]
3 years ago
5

Suppose a random sample of 100 observations from a binomial population gives a value of pˆ = .63 and you wish to test the null h

ypothesis that the population parameter p is equal to .70 against the alternative hypothesis that p is less than .70.
a) nothing that p = .63, what does your intuition tell you? does the value of p appear to contradict the null hypothesis?
b) use the large-sample z-test to test H0: p = .70 against the alternative hypothesis, Ha: p < 70. Use a = .05. How do the test results compare with your intuitive decision from part A?
c) find and interpret the observed significance level of the test you conducted in part B.
Mathematics
1 answer:
irakobra [83]3 years ago
4 0

Answer:

We conclude that the population proportion is equal to 0.70.

Step-by-step explanation:

We are given that a random sample of 100 observations from a binomial population gives a value of pˆ = 0.63 and you wish to test the null hypothesis that the population parameter p is equal to 0.70 against the alternative hypothesis that p is less than 0.70.

Let p = <u><em>population proportion.</em></u>

(1) The intuition tells us that the population parameter p may be less than 0.70 as the sample proportion comes out to be less than 0.70 and also the sample is large enough.

(2) So, Null Hypothesis, H_0 : p = 0.70      {means that the population proportion is equal to 0.70}

Alternate Hypothesis, H_A : p < 0.70      {means that the population proportion is less than 0.70}

The test statistics that would be used here <u>One-sample z-test</u> for proportions;

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

where, \hat p = sample proportion = 0.63

            n = sample of observations = 100

So, <u><em>the test statistics</em></u>  =  \frac{0.63-0.70}{\sqrt{\frac{0.70(1-0.70)}{100} } }

                                     =  -1.528

The value of z-test statistics is -1.528.

<u>Now at 0.05 level of significance, the z table gives a critical value of -1.645 for the left-tailed test.</u>

Since our test statistics is more than the critical value of z as -1.528 > -1.645, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u><em>we fail to reject our null hypothesis</em></u>.

Therefore, we conclude that the population proportion is equal to 0.70.

(c) The observed level of significance in part B is 0.05 on the basis of which we find our critical value of z.

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