For this hypothesis problem, must include: step 1: set a set of hypothesesstep 2: determine rejection rule with given astep 3: compute test statistic (x2-value)step 4: make judgment (reject H0 or not) Please help. How do I solve this #2. The manager of Apple Green Golf wishes to study the number of rounds of golf played per weekday. He gathered the following sample information. At the 0.05 significance level, is there a difference in the number of rounds played by day of the week (or the same)? (Perform Goodness-of-Fit test)| Day | Rounds ||-----------|--------|| Monday | 124 || Tuesday | 74 || Wednesday | 104 || Thursday | 98 || Friday | 120 | To determine if there is a statistical difference in the number of golf rounds played per weekday, a Goodness-of-Fit test should be performed using the above data at a 0.05 significance level.

The manager of Apple Green Golf wishes to study the number of rounds of golf played per week day. He gathered the following sample information. Day Rounds Monday 124 Tuesday 74 Wednesday 104 Thursday 98 Friday 120 Total 520 At the 0.05 significance level, we need to perform Chi-square Goodness-of-fit test to check if there is there a difference in the number of rounds played by day of the week (or the same). The null and alternative hypothesis for the Chi-square Goodness-of-fit test are as follows, H0 : The number of rounds played by day of the week are same. Vs H1 : The number of rounds played by day of the week are different. For Chi-square test we need to find the expected frequencies (Ei) for each day as, Ei=Nk, where N = Total number of frequencies and k = number of categories/groups. This gives, Ei=5205=104. And hence the table is given by, Day Rounds (Oi) Expected (Ei) Monday 124 104 Tuesday 74 104 Wednesday 104 104 Thursday 98 104 Friday 120 104 Total 520 520 The test statistic for test is given by the formula, Where degrees of freedom (df) = k-1 = 5-1 = 4. The p-value in this context is given by, The expected frequencies can be found using the formula given in step 1. Consider the following table of calculations to get the value of test statistic. Day Rounds (Oi) Expected (Ei) (Oi-Ei)² (Oi-Ei)²/Ei Monday 124 104 400 3.8462 Tuesday 74 104 900 8.6538 Wednesday 104 104 0 0.0000 Thursday 98 104 36 0.3462 Friday 120 104 256 2.4615 Total 520 520 15.3077 The sum of last column is the required value of test statistic. This gives, χ02=15.3077. The p-value for the test is given as follows, p-value=Pχdf2>χ02=Pχ42>15.0377=0.0041 The p-value can be calculated using the excel function, "=CHISQ.DIST.RT(χ02,df)"To make decision based on the p-value we need to know the significance level (α). The decision rule based on p-value is given by, "If p-value ≤ α then we reject H0 otherwise we fail to reject H0." We are given the significance level, α=0.05. Decision: Since p-value = 0.0041 < α = 0.05 we reject null hypothesis (H0). Conclusion: There is sufficient evidence to support the claim that, the number of rounds played by day of the week are different.