On an intramural softball team, the proportion of hits to at bats for the entire team during the last season was 30% of 300 attempts. Estimate the true proportion of hits using a 90% CI. The answers need to be proportions (not percents) and rounded to the nearest hundredth (two (2) decimal places) to be counted as correct. (For example, if my CI is (0.1002, 0.2159) then they need to be input as 0.10 and 0.22 to be correct. These are not the answers to this question :-) )

The lower bound is and the upper bound is

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JeanaShupp JeanaShupp · Answer 1 · 2019-12-25T05:57:32+01:00

Answer: The lower bound is 0.26 and the upper bound is 0.34.

Step-by-step explanation:

Formula to find the confidence interval for population proportion (p) is given by :_

[tex]\hat{p}\pm z^* \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

, where n= sample size

z* = Critical value. (two-tailed)

[tex]\hat{p}[/tex] = Sample proportion.

Let p be the true population proportion of hits to at bats for the entire team during the last season.

As per given , we have

n= 300

[tex]\hat{p}=0.30[/tex]

By z-table , the critical value for 90% confidence interval : z* = 1.645

Now , 90% confidence interval for the proportion of hits to at bats for the entire team during the last season:

[tex]0.30\pm (1.645) \sqrt{\dfrac{0.30(1-0.30)}{300}}[/tex]

[tex]0.30\pm (1.645) \sqrt{0.0007}[/tex]

[tex]0.30\pm (1.645) (0.0264575131106)[/tex]

[tex]\approx0.30\pm0.0435[/tex]

[tex]=(0.30-0.0435,\ 0.30+0.0435)\\\\=(0.2565,\ 0.3435)\approx(0.26,\ 0.34)[/tex]

The lower bound is 0.26 and the upper bound is 0.34.