Question

A representative sample of 35 customers at bagel shop were surveyed and ask if they prefer sandwiches made on plain, everything, or sesame bagels. Each participant could only select their one most favorite bagel type. In that sample, 16 selected plain, 12 selected everything, and 7 selected sesame. Conduct a chi-square goodness-of-fit test to determine if there is evidence that the proportions of people in the population who would choose plain, everything, and sesame bagels are not all equal. Use Minitab Express and remember to include all relevant output. [25 points]

Step 1: State hypotheses and check assumptions

Step 2: Compute the test statistic

Step 3: Determine the p-value

Step 4: Decide to reject or fail to reject the null hypothesis

Step 5: State a real-world conclusion

Answer 1

## We have to use MINITAB and do chi square goodness of fit :

here total observation = 35

MINITAB  c1 : Sandwich and c2 = count

Stat > table > chi square goodness of fit > observed count select count ( c2 )

> categorical count select count (c2)

## Step 1: State hypotheses and check assumptions

Answer :

Ho : all proportion are equal for sandwiches vs

H1 : not all proportion are equal for sandwiches

we can use chi square goodness of fit because we have one variable and it is categorical :

it is necessary to do chi square test ( assumption )

Step 2: Compute the test statistic

Answer : chi square =  3.48571

Step 3: Determine the p-value

Answer : 0.175

Step 4: Decide to reject or fail to reject the null hypothesis

Decision : we reject Ho if p value is less than α value using p value approach here p value is > α

value we fail to reject Ho at given level of significance .

Step 5: State a real-world conclusion

There is Insufficient evidence that the proportions of people in the population who would choose plain , everything and sesame bagels are not equal .

( ie proportions are equal )

## Output of the chi square goodness of fit :

MTB > TChiSquare;
SUBC> Observed 'count';
SUBC> GBar;
SUBC> GChiSQ;
SUBC> Pareto;
SUBC> RTable.

Chi-Square Goodness-of-Fit Test for Observed Counts in Variable: count

Test Contribution to
Category Observed Proportion Expected Chi-Sq
1 16 0.333333 11.6667 1.60952
2 12 0.333333 11.6667 0.00952
3 7 0.333333 11.6667 1.86667

N DF Chi-Sq P-Value
35 2 3.48571 0.175