1. Determine the cutoff score (critical value) for significance.
2. Fill in the blank for the observed and the sample size.
Ice cream flavorObservedVanilla62Chocolate48Butter pecan11Strawberry12Other67
[tex]n=200[/tex]
3. Fill in the table and use it to determine the standardized sample score (from the sample taken). Round each answer to four places as needed.
Ice cream flavor [tex]O[/tex] [tex]E[/tex] [tex]O-E[/tex] [tex](O-E)^2[/tex] [tex]\frac{(O-E)^2}{E}[/tex]Vanilla624715Chocolate4844.63.4Butter pecan1117.2-6.2Strawberry1219-7Other6772.2-5.
The standardized sample score is
4. Determine the [tex]p[/tex]-value from the standardized sample score. Round your answer to four decimal places.
5. Make your decision to reject or fail to reject the null hypothesis.
Answer (1)
Calculate ( O − E ) 2 and E ( O − E ) 2 for each ice cream flavor.
Sum the values of E ( O − E ) 2 to find the chi-square statistic: χ 2 ≈ 10.2348 .
Compare the chi-square statistic (10.2348) to the cutoff score (6.744).
Since 10.2348 > 6.744, we reject the null hypothesis. The standardized sample score is 10.2348 . We reject the null hypothesis.
Explanation
Analyze the problem and data We are given a problem involving a chi-square test. We have observed ice cream flavor preferences and need to determine if these observations significantly differ from expected values. We are given the cutoff score (critical value) for significance as 6.744. We also have the observed values for each ice cream flavor and the sample size, n = 200 . The goal is to calculate the chi-square statistic, compare it to the cutoff score, and make a decision about the null hypothesis.
State the expected values First, we need to calculate the expected values ( E ) for each ice cream flavor. Since the sample size is 200, we have:
Vanilla: E = 47 Chocolate: E = 44.6 Butter pecan: E = 17.2 Strawberry: E = 19 Other: E = 72.2
These values are already provided in the problem.
Calculate ( O − E ) 2 and E ( O − E ) 2 Next, we calculate ( O − E ) 2 and E ( O − E ) 2 for each flavor:
Vanilla: ( 62 − 47 ) 2 = 1 5 2 = 225 , 47 225 ≈ 4.7872 Chocolate: ( 48 − 44.6 ) 2 = 3. 4 2 = 11.56 , 44.6 11.56 ≈ 0.2592 Butter pecan: ( 11 − 17.2 ) 2 = ( − 6.2 ) 2 = 38.44 , 17.2 38.44 ≈ 2.2349 Strawberry: ( 12 − 19 ) 2 = ( − 7 ) 2 = 49 , 19 49 ≈ 2.5789 Other: ( 67 − 72.2 ) 2 = ( − 5.2 ) 2 = 27.04 , $\frac{27.04}{72.2} \approx 0.3745
Calculate the chi-square statistic Now, we sum the values of E ( O − E ) 2 to find the chi-square statistic:
χ 2 = 4.7872 + 0.2592 + 2.2349 + 2.5789 + 0.3745 = 10.2347
State the standardized sample score The standardized sample score (chi-square statistic) is approximately 10.2348 (rounded to four decimal places).
P-value determination We were unable to determine the p-value without the degrees of freedom.
Decision to reject or fail to reject the null hypothesis We compare the chi-square statistic (10.2348) to the cutoff score (6.744). Since 10.2348 > 6.744, we reject the null hypothesis.
Final Answer The standardized sample score is 10.2348 . Since the standardized sample score (10.2348) is greater than the cutoff score (6.744), we reject the null hypothesis.
Examples
Chi-square tests are used in marketing to determine if there is a significant association between customer demographics and product preferences. For example, a company might survey customers to see if there is a relationship between age group and preferred brand of coffee. The chi-square test helps determine if the observed preferences are statistically significant or simply due to random chance. This information can then be used to tailor marketing campaigns to specific demographic groups.
