The following table shows results from a general survey conducted by a dating web site from a random sample of 100 people on the type of dating activity and whether a person's enjoyed their date.Results of Date Type of Activity Good Date Bad Date Neutral Row TotalIndoor Activity 21 23 9 53Outdoor Activity 30 12 5 47Column Total 51 35 14 100Use the chi-square test to determine that the type of dating activity and results of a date are independent at the 0.05 level of significance.(a)What is the level of significance? State the null and alternate hypotheses.H0: The type of dating activity and date results are independent.H1: The type of dating activity and date results are not independent.H0: The type of dating activity and date results are not independent.H1: The type of dating activity and date results are not independent. H0: The type of dating activity and date results are independent.H1: The type of dating activity and date results are independent.H0: The type of dating activity and date results are not independent.H1: The type of dating activity and date results are independent.(b)Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)Are all the expected frequencies greater than 5?YesNo What sampling distribution will you use?binomialStudent's t chi-squarenormaluniformWhat are the degrees of freedom? (c)Find or estimate the P-value of the sample test statistic.P-value > 0.1000.050 < P-value < 0.100 0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005(d)Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence?Since the P-value > 𝛼, we reject the null hypothesis.Since the P-value ≤ 𝛼, we reject the null hypothesis. Since the P-value > 𝛼, we fail to reject the null hypothesis.Since the P-value ≤ 𝛼, we fail to reject the null hypothesis.(e)Interpret your conclusion in the context of the application.At the 5% level of significance, there is sufficient evidence to conclude that the type of dating activity and results of a date are not independent.At the 5% level of significance, there is insufficient evidence to conclude that the type of dating activity and results of a date are not independent.
Question
The following table shows results from a general survey conducted by a dating web site from a random sample of 100 people on the type of dating activity and whether a person's enjoyed their date.Results of Date Type of Activity Good Date Bad Date Neutral Row TotalIndoor Activity 21 23 9 53Outdoor Activity 30 12 5 47Column Total 51 35 14 100Use the chi-square test to determine that the type of dating activity and results of a date are independent at the 0.05 level of significance.(a)What is the level of significance? State the null and alternate hypotheses.H0: The type of dating activity and date results are independent.H1: The type of dating activity and date results are not independent.H0: The type of dating activity and date results are not independent.H1: The type of dating activity and date results are not independent. H0: The type of dating activity and date results are independent.H1: The type of dating activity and date results are independent.H0: The type of dating activity and date results are not independent.H1: The type of dating activity and date results are independent.(b)Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)Are all the expected frequencies greater than 5?YesNo What sampling distribution will you use?binomialStudent's t chi-squarenormaluniformWhat are the degrees of freedom? (c)Find or estimate the P-value of the sample test statistic.P-value > 0.1000.050 < P-value < 0.100 0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005(d)Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence?Since the P-value > 𝛼, we reject the null hypothesis.Since the P-value ≤ 𝛼, we reject the null hypothesis. Since the P-value > 𝛼, we fail to reject the null hypothesis.Since the P-value ≤ 𝛼, we fail to reject the null hypothesis.(e)Interpret your conclusion in the context of the application.At the 5% level of significance, there is sufficient evidence to conclude that the type of dating activity and results of a date are not independent.At the 5% level of significance, there is insufficient evidence to conclude that the type of dating activity and results of a date are not independent.
Solution
(a) The level of significance is 0.05. The null hypothesis (H0) is that the type of dating activity and date results are independent. The alternate hypothesis (H1) is that the type of dating activity and date results are not independent.
(b) To find the value of the chi-square statistic for the sample, we first need to calculate the expected frequencies for each cell in the table. The expected frequency for each cell is calculated as (row total * column total) / grand total.
For example, the expected frequency for 'Good Date' and 'Indoor Activity' is (53*51)/100 = 27.03. We do this for all cells and then calculate the chi-square statistic using the formula: Χ² = Σ [ (O-E)² / E ] where O is the observed frequency and E is the expected frequency.
If all the expected frequencies are greater than 5, we can use the chi-square distribution. The degrees of freedom for the chi-square test is calculated as (number of rows - 1) * (number of columns - 1) = (2-1)*(3-1) = 2.
(c) The P-value is the probability that, if the null hypothesis were true, we would observe a more extreme test statistic in the direction of the alternative hypothesis than we did. We can find this value from the chi-square distribution table using the calculated chi-square statistic and degrees of freedom.
(d) If the P-value is less than or equal to the level of significance (0.05), we reject the null hypothesis. If the P-value is greater than the level of significance, we fail to reject the null hypothesis.
(e) Depending on the result of the test, we can conclude whether there is sufficient or insufficient evidence at the 5% level of significance to say that the type of dating activity and results of a date are not independent.
Similar Questions
A table classifying a random sample of students by course and whether or not they are "binge drinkers" is given below.Science (BSc, BBiomed, BEng) Arts (BA, BMusic) Total"binge drinker" 68 63 131not "binge drinker" 355 423 778Total 423 486 909When testing the null hypothesis that there is no association between gender and "binge drinking", which ONE of the following statements is FALSE?Group of answer choicesThe assumptions for the chi-squared test are invalid.A suitable test statistic has a value of 1.33A suitable test statistic has a value of 1.776The p-value for the test is 0.183Based on this data, there is insufficient evidence to suggest that science students are more likely than arts students to be "binge drinkers".
What does a chi-square test compare?Question 4Answera.the frequency of incorrect responses for each variableb.correct and incorrect reaction timesc.the observed and total frequencies for each individual participantd.the observed and expected frequencies for each cell
3. The Chi Square test is applicable for a given data, if*
A hypothesis was tested at an α = 0.05 significance level. Review the table below and complete the sentence that follows:Table 1 ScoreSum of squaresdfMean squareFSig. (p)Between groups11.57125.7867.8180.02Within groups154.2346240.740Total165.805626Based on the information presented in Table 1, the decision is taken to __________.Group of answer choicesReject the null hypothesisAccept the null hypothesisReject the alternative hypothesisAccept the alternative hypothesis
Which of the following statements about the Chi-square analysis is true? Categorical data from questions about sex, education, or other nominal variables cannot be tested with this statistic. The expected frequencies in Chi-square cannot be calculated theoretically. The larger the Chi-square, the less likely it is that the two variables that are measured are related. When the differences between observed and expected frequencies are large, one must accept the null hypothesis. Chi-square analysis compares the observed frequencies of responses with the expected frequencies.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.