Answer to Question #208828 in Statistics and Probability for KAYLA

Question

In order for a candy company to claim that a bridge mix is mostly chocolate stars, a proportion of at least 0.8 of the packages must contain 3 ounces or more of chocolate stars. Quality control tests a random sample of 50 packages to determine if the proportion is less than 0.8 at a significance level of 0.05. Use Sheet 4 of the Excel file to calculate p^

, the z

-score, and the p

-value

Answer 1

Total problem with data

In order for a candy company to claim that a bridge mix is mostly chocolate stars, a proportion of at least 0.8 of the packages must contain 3 ounces or more of chocolate stars. Quality control tests a random sample of 50 packages to determine if the proportion is less than 0.8 at a significance level of 0.5. Use Sheet 5 of the Excel file to calculate p^ , the z -score, and the p-value. p^ = _______ z = ________ p = ________ Since the p-value is______ than the significance level 0.5, the null hypothesis_______ ________evidence exists that the proportion of packages with 3 ounces or more of chocolate stars is less than 0.8 .

3.273525

3.725058

3.193881

3.380596

3.849062

2.688789

4.521501

2.876926

3.202865

3.760863

3.648738

3.155042

3.047772

2.331491

2.434267

3.316263

3.335704

3.123784

1.983015

4.026382

2.819852

3.020835

3.963952

3.434813

2.530151

3.501124

2.723868

3.734019

3.074854

2.94544

2.963046

3.301251

3.806637

2.49562

2.990224

2.548798

2.849587

3.55404

3.565047

3.383945

2.895875

3.960068

3.744978

4.034873

3.394484

4.637022

3.085

3.945715

2.590541

3.59302

Range data and count the number of samples more than 3 (n).

N=50

n=34

Proportion:

$\hat{p}=\frac{n}{N}= \frac{34}{50}=0.68 \\ H_0: p≥0.8\\ H_1: p<0.8 \\ Z = \frac{\hat{p}-p}{\sqrt{ \frac{p(1-p)}{N} }} \\ = \frac{0.68-0.8}{ \sqrt{ \frac{0.8(1-0.8)}{50} } } \\ = -2.121 \\ α=0.05$

Reject H₀ if $Z< -Z_{critical}$

$Z_{critical}$ for one tailed (Left) = -1.6449

$Z=-2.121 < Z_{critical}$

Reject H₀.

P(Z < -2.121) = 0.0169

p-value = 0.0169

Since the p-value is less than significance level of 0.05 the null hypothesis is rejected and $Z=-2.121 < Z_{critical}$ . There is enough evidence to conclude that the proportion of packages with 3 ounces or more of chocolate stars is less than 0.8 at 0.05 significance level.