Investigate two varieties of Kisses: Hershey Kisses with Almonds and Regular Her

Question

Investigate two varieties of Kisses: Hershey Kisses with Almonds and Regular Hershey Kisses.

Means

a) Run a test to determine if the meanweights are different. Show all your work by handusing a test statistic / critical valueapproach. Use alpha = .05

b) Run the above test using a p-value / level of significanceapproach. Get an exact p-value: Example t-test using Excel

c)What are the populations?

d)What are the population parameters?

e)What are the sample point estimates?

f)What is your conclusion about the difference in the means? Variances

g) Run a test to compare the two population variances. Determine if these variancesare different. Show all your work by hand using a test statistic / critical valueapproach. Use alpha = .05

h) Run the above test using a p-value / level of significanceapproach. Get an exact p-value: Example F-Test using Excel

i) What are the population parameters?

j)What are the sample point estimates?

k)What is your conclusion about the difference in the variances?

Kiss Data

Almonds Regular 4.5 4.77 4.48 4.68 4.66 4.83 4.57 4.89 4.68 4.85 4.71 4.37 4.67 4.9 4.6 4.67 4.64 4.67 4.65 4.53 4.79 4.89 4.73 4.59 4.61 4.56 4.66 4.65 4.54 4.67 4.76 4.69 4.68 4.39 4.71 4.7 4.61 4.76 4.62 4.72 4.65 4.69

Explanation / Answer

a) Run a test to determine if the mean weights are different. Show all your work by hand using a test statistic / critical value approach. Use alpha = .05

Solution:

Here, we have to use the two sample t test for the population means. The null and alternative hypothesis for this test is given as below:

H0: µ1 = µ2 versus Ha: µ1 µ2

This is two tailed test.

We are given alpha = 0.05

Test statistic = t= (X1bar – X2bar) / sqrt[(S1^2/N2)+(S2^2/N2)]

From the given data we have

X1bar = 4.643809524

X2bar = 4.689047619

S1^2 = 0.006174762

S2^2 = 0.021469048

N1 = N2 = 21

Degrees of freedom = 21 + 21 – 2 = 40

Test statistic = t = (4.643809524 - 4.689047619) / sqrt[(0.006174762/21)+( 0.021469048/21)]

Test statistic = t = -1.246852286

Critical value = t = -2.02107537 and 2.02107537

Here, absolute value of test statistic < Critical value

So, we do not reject the null hypothesis that there is no significant difference in the mean weights of Hershey Kisses with Almonds and Regular Hershey Kisses.

b) Run the above test using a p-value / level of significance approach. Get an exact p-value: Example t-test using Excel

Solution:

The above test by using excel is given as below:

t-Test: Two-Sample Assuming Equal Variances

Almonds

Regular

Mean

4.643809524

4.689047619

Variance

0.006174762

0.021469048

Observations

21

21

Pooled Variance

0.013821905

Hypothesized Mean Difference

0

df

40

t Stat

-1.246852304

P(T<=t) one-tail

0.10985179

t Critical one-tail

1.683851014

P(T<=t) two-tail

0.21970358

t Critical two-tail

2.02107537

P-value = 0.2197

Alpha value = 0.05

P-value > Alpha value, so we do not reject the null hypothesis that there is no significant difference in the mean weights of Hershey Kisses with Almonds and Regular Hershey Kisses.

c) What are the populations?

Solution:

The two populations for this test are considered as the weights of all Hershey Kisses with Almonds and Regular Hershey Kisses.

d) What are the population parameters?

Solution:

The average weights of Hershey Kisses with Almonds and Regular Hershey Kisses are the population parameters for this test.

e) What are the sample point estimates?

Solution:

The sample point estimates for this test is given as the sample means and sample variances.

t-Test: Two-Sample Assuming Equal Variances

Almonds

Regular

Mean

4.643809524

4.689047619

Variance

0.006174762

0.021469048

Observations

21

21

Pooled Variance

0.013821905

Hypothesized Mean Difference

0

df

40

t Stat

-1.246852304

P(T<=t) one-tail

0.10985179

t Critical one-tail

1.683851014

P(T<=t) two-tail

0.21970358

t Critical two-tail

2.02107537

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