Question

You intend to estimate a population proportion with a confidence interval. The data suggests that the normal distribution is a reasonable approximation for the binomial distribution in this case.

Find the critical value that corresponds to a confidence level of 90%.

(Report answer accurate to three decimal places with appropriate rounding.)

Step 1

If *X *is a binomial random variable, then *X*~*B *(*n, p*) where *n *is the number of trials and *p *is the probability of success. The proportion is obtained as the fraction of *X*, the random variable for the number of success with the number of trials (*n*) or the sample size.

The proportion is denoted by, P^(read as P-hat). That is,

P^=X/n.

The normal distribution can be used to approximate the binomial, when *n* is large and *p *is not close to zero or one.

That is, *X*~*N *(*np, sqrt*(*npq*)).

If the random variable, the mean and the standard deviation is divided with *n, *a normal distribution of proportions with P^ (called estimated proportion) as the random variable is obtained. That is,

Step 2

The approximation can be used only if the number of success, np^ and the number of failures, n(1–p^) are both greater than 5.

Step 3

**Critical value:**

The confidence level is 90%.

Thus, the level of significance is *α* = 0.10.

*Critical value for left tailed test:*

The *z*-critical value for a left tailed test at *α* = 0.10 is z0.10(=zα).

The corresponding value is obtained using the EXCEL formula,

“=NORM.INV(0.10,0,1)”.

Thus, the critical value for a left tailed test at 90% confidence level is **–1.280**.

*Critical value for right tailed test:*

The *z*-critical value for a right tailed test at *α* = 0.10 is z0.90(=z1-α).

The corresponding value is obtained using the EXCEL formula,

“=NORM.INV(0.90,0,1)”.

T...

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