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1) According to Mars, Inc., 20% of all M&Ms produced are blue. One bag of 50 M&Ms represents the sample for this problem. The sample data can be used to perform a two-sided hypothesis test to test whether 20% of all M&Ms are blue.
In one bag of 50 M&Ms, there are 18 blue M&Ms. Use this data to test whether 20% of all M&Ms are blue.
Based on the given information and using the appropriate formula, calculate the test statistic for this hypothesis test. Round your answer to three decimal places. Enter the numeric value of the test statistic in the space below:
2) According to Mars, Inc., 20% of all M&Ms produced are blue. One bag of 50 M&Ms represents the sample for this problem. The sample data can be used to perform a two-sided hypothesis test to test whether 20% of all M&Ms are blue.
In one bag of 50 M&Ms, there are 14 blue M&Ms. Use this data to test whether 20% of all M&Ms are blue.
Based on the given information and using the appropriate formula, calculate the test statistic for this hypothesis test. Round your answer to three decimal places. Enter the numeric value of the test statistic in the space below:
3) You are given the following null and alternate hypotheses:
H : p = .55
H : p < .55
The test statistic for this hypothesis test is z = -0.35.
For this problem, use the provided Standard Normal Table (Z table). Determine the p-value for this test. Enter the p-value in the space below as a decimal rounded to four decimal places:
4) You are given the following null and alternate hypotheses:
H : p = .55
H : p < .55
The test statistic for this hypothesis test is z = -2.46.
For this problem, use the provided Standard Normal Table (Z table). Determine the p-value for this test. Enter the p-value in the space below as a decimal rounded to four decimal places:
5) You are given the following null and alternate hypotheses:
H : p = .55
H : p < .55
The test statistic for this hypothesis test is z = -1.89.
For this problem, use the provided Standard Normal Table (Z table). Determine the p-value for this test. Enter the p-value in the space below as a decimal rounded to four decimal places:

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Answer:

Check the explanation

Step-by-step explanation:

Given:

Claim: According to Mars, Inc., 20% of all M&Ms produced are blue.

that is: p = proportion of M&Ms are blue = 0.20

Sample size = n = 50

x = Number of M&Ms are blue = 14

Step 1) State H0 and H1:

H_{0}: p = 0.20 Vs H_{1}: p \neq 0.20

Step 2) Test statistic:

P-p px(1-p)

where

[tex]\hat{p}=\frac{x}{n} = \frac{14}{50}[/tex] = 0.28

thus

P-p px(1-p)

z= [tex]\frac{0.28 - 0.20}{\sqrt{\frac{0.20 \times (1-0.20)}{50}}}[/tex]

z=[tex]\frac{0.08 }{\sqrt{\frac{0.20 \times 0.80}{50}}}[/tex]

z= [tex]\frac{0.08 }{\sqrt{ 0.0032 }}[/tex]

z= [tex]\frac{0.08 }{ 0.0565685 }[/tex]

z= 1.41

Step 3) Find z critical value:

Since level of significance is not given , we use level of significance = g= 0,05

Look in z table for area = 0.0250 or its closest area and find z value

Area 0.0250 corresponds to -1.9 and 0.06

thus z critical value = -1.96

Since this is two tailed test, we have two z critical values: ( -1.96 , 1.96)

Step 4) Decision Rule:

Reject null hypothesis ,if z  test statistic value < z critical value =-1.96 or z  test statistic value > z critical value = 1.96 , otherwise we fail to reject H0.

That is rejection region is: z < -1.96 or z > 1.96.

Since z  test statistic value = z= 1.41 is between z critical values: ( -1.96 , 1.96), that is z  test statistic value = z= 1.41 does not fall in the rejection region, we fail to reject H0.

Step 5) Conclusion:

At 0.05 level of significance, we do not have sufficient evidence to reject the claim that: 20% of all M&Ms produced are blue.

that is: Mars, Inc. claim that 20% of all M&Ms produced are blue is true.

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