Question

1.Compute the population proportion interval estimate given n, p, and the confidence level.

a. n = 300 and p = 0.40 with 95% confidence

b. n = 500 and p = 0.23 with 95% confidence

c. n = 420 and p = 0.61 with 90% confidence

d. n = 670 and p = 0.54 with 99% confidence

e. n = 710 and p = 0.63 with 99% confidence

2.Ismail conducted a poll survey in which 320 and 600 randomly selected voters indicated their preference for a certain candidate. Using 95% confidence level, what is the interval containing the true population proportion p of voters who prefer the candidate?

3.A random sample of 180 users of a new kind of organic shampoo yielded p = 0.64. Use a = 0.05.

a. What is the point estimate of p?

b. What is the interval estimate of p?

4.A random sample of 156 packets of cupcakes yield p = 0.76. Construct a 90% confidence interval for p.

Answer 1

1.

(a) Given, n=300, p=0.4

    q=1-0.4=0.6

 Here $Z_{\dfrac{\alpha}{2}}=Z_{0.025}$ =1.96

 Confidence level $=p\pm Z_{\frac{\alpha}{2}}\sqrt{\dfrac{pq}{n}}$

           $=0.4\pm 1.96 \sqrt{\dfrac{0.4\times 0.6}{300}}$

$=0.4\pm 0.0554 =(0.3446,0.4554)$

(b) Given, n=500, p=0.23

    q=1-0.23=0.77

 Here $Z_{\dfrac{\alpha}{2}}=Z_{0.025}$ =1.96

 Confidence level $=p\pm Z_{\frac{\alpha}{2}}\sqrt{\dfrac{pq}{n}}$

           $=0.23\pm 1.96 \sqrt{\dfrac{0.23\times 0.77}{500}}$

$=0.23\pm 0.0368 =(0.1931,0.2668)$

(c) Given, n=420, p=0.61

    q=1-0.61=0.39

 Here $Z_{\dfrac{\alpha}{2}}=Z_{0.05}$ =1.645

 Confidence level $=p\pm Z_{\frac{\alpha}{2}}\sqrt{\dfrac{pq}{n}}$

           $=0.61\pm 1.645 \sqrt{\dfrac{0.61\times 0.39}{420}}$

$=0.61\pm 0.0.03915 =(0.5708,0.64915)$

(d) Given, n=670, p=0.54

    q=1-0.54=0.46

 Here $Z_{\dfrac{\alpha}{2}}=Z_{0.005}$ =2.342

 Confidence level $=p\pm Z_{\frac{\alpha}{2}}\sqrt{\dfrac{pq}{n}}$

           $=0.54\pm 2.342 \sqrt{\dfrac{0.54\times 0.46}{670}}$

$=0.54\pm 0.01925 =(0.5207,0.55925)$

(e) Given, n=710, p=0.63

    q=1-0.63=0.37

 Here $Z_{\dfrac{\alpha}{2}}=Z_{0.005}$ =2.342

 Confidence level $=p\pm Z_{\frac{\alpha}{2}}\sqrt{\dfrac{pq}{n}}$

           $=0.63\pm 2.342 \sqrt{\dfrac{0.63\times 0.37}{710}}$

$=0.63\pm 0.042435 =(0.58756,0.67243)$

(2). n=600, p=\dfrac{320}{600}=0.53

q=1-p=1-0.53=0.47

$Z_{\frac{\alpha}{2}}=Z_{\frac{0.05}{2}}=z_{0.025}=1.96$

 Confidence level $=p\pm Z_{\frac{\alpha}{2}}\sqrt{\dfrac{pq}{n}}$

$=0.53\pm 1.96\sqrt{\dfrac{0.53\times 0.47}{600}}$

$=0.53\pm 0.0399=(0.49006,0.56993)$

(3). n=180, p=0.64

q=1-p=1-0.64=0.36

$Z_{\frac{\alpha}{2}}=Z_{\frac{0.05}{2}}=z_{0.025}=1.96$

(i) Point estimate of $p =\dfrac{p}{n}=\dfrac{0.64}{180}=0.00355$

(ii) Interval estimates

 Confidence level $=p\pm Z_{\frac{\alpha}{2}}\sqrt{\dfrac{pq}{n}}$

$=0.64\pm 1.96\sqrt{\dfrac{0.64\times 0.36}{180}}$

$=0.64\pm 0.07012=(0.56987,0.71012)$

(4). Given, n=156, p=0.76

    q=1-0.76=0.24

 Here $Z_{\dfrac{\alpha}{2}}=Z_{0.05}$ =1.645

 Confidence level $=p\pm Z_{\frac{\alpha}{2}}\sqrt{\dfrac{pq}{n}}$

           $=0.76\pm 1.645 \sqrt{\dfrac{0.76\times 0.24}{156}}$

$=0.76\pm 0.05624 =(0.70375,0.81624)$