is race a factor in the number of times a mon...
is race a factor in the number of times a month a person eats out? the table below shows data that was collected. assume that all distributions are normal, the four population standard deviations are all the same, and the data was collected independently and randomly. use a level of significance of $alpha = 0.05$ 1. for this study, we should use one - way anova 2. your friend helped you with the null and alternative hypotheses $h_0:mu_1=mu_2=mu_3=mu_4$ $h_1:$ at least one of the means is different from the others. 3. the test - statistic for this data = (please show your answer to 3 decimal places.) 4. the p - value for this sample = (please show your answer to 4 decimal places.) 5. the p - value is greater than alpha $alpha$ 6. based on this, we should fail to reject the null hypothesis 7. as such, the final conclusion is that... there is sufficient evidence to support the claim that race is a factor in the number of times per month people eat out.
Answer
# Explanation:
## Step1: Calculate group means and grand mean
Let $n_1 = 9$, $n_2=8$, $n_3 = 7$, $n_4=7$.
For Asian group: $\bar{x}_1=\frac{7 + 3+15+15+6+8+12+8+12}{9}=\frac{86}{9}\approx9.556$
For White group: $\bar{x}_2=\frac{6 + 12+8+10+12+2+1+5}{8}=\frac{56}{8} = 7$
For Black group: $\bar{x}_3=\frac{4+3+5+5+15+6+9}{7}=\frac{47}{7}\approx6.714$
For Hispanic group: $\bar{x}_4=\frac{4+15+13+14+3+0+11}{7}=\frac{50}{7}\approx7.143$
The grand - mean $\bar{\bar{x}}=\frac{86 + 56+47+50}{9 + 8+7+7}=\frac{239}{31}\approx7.71$
## Step2: Calculate sum of squares between groups (SSB)
$SSB=\sum_{i = 1}^{k}n_i(\bar{x}_i-\bar{\bar{x}})^2$
$SSB=9\times(9.556 - 7.71)^2+8\times(7 - 7.71)^2+7\times(6.714 - 7.71)^2+7\times(7.143 - 7.71)^2$
$SSB=9\times(1.846)^2+8\times(- 0.71)^2+7\times(-0.996)^2+7\times(-0.567)^2$
$SSB=9\times3.407+8\times0.504+7\times0.992+7\times0.322$
$SSB = 30.663+4.032+6.944+2.254$
$SSB=43.893$
## Step3: Calculate sum of squares within groups (SSW)
For Asian group: $SSW_1=\sum_{j = 1}^{n_1}(x_{1j}-\bar{x}_1)^2=(7 - 9.556)^2+(3 - 9.556)^2+\cdots+(12 - 9.556)^2$
$SSW_1 = 79.556$
For White group: $SSW_2=\sum_{j = 1}^{n_2}(x_{2j}-\bar{x}_2)^2=(6 - 7)^2+(12 - 7)^2+\cdots+(5 - 7)^2$
$SSW_2=50$
For Black group: $SSW_3=\sum_{j = 1}^{n_3}(x_{3j}-\bar{x}_3)^2=(4 - 6.714)^2+(3 - 6.714)^2+\cdots+(9 - 6.714)^2$
$SSW_3=46.857$
For Hispanic group: $SSW_4=\sum_{j = 1}^{n_4}(x_{4j}-\bar{x}_4)^2=(4 - 7.143)^2+(15 - 7.143)^2+\cdots+(11 - 7.143)^2$
$SSW_4=75.714$
$SSW=SSW_1+SSW_2+SSW_3+SSW_4=79.556 + 50+46.857+75.714=252.127$
## Step4: Calculate degrees of freedom
Degrees of freedom between groups $df_{B}=k - 1=4 - 1 = 3$
Degrees of freedom within groups $df_{W}=N - k=31 - 4=27$
## Step5: Calculate mean square between groups (MSB) and mean square within groups (MSW)
$MSB=\frac{SSB}{df_{B}}=\frac{43.893}{3}\approx14.631$
$MSW=\frac{SSW}{df_{W}}=\frac{252.127}{27}\approx9.338$
## Step6: Calculate the test - statistic
The test - statistic $F=\frac{MSB}{MSW}=\frac{14.631}{9.338}\approx1.567$
## Step7: Calculate the p - value
Using an F - distribution table or statistical software with $df_{B}=3$ and $df_{W}=27$ and $F = 1.567$, the p - value is approximately $0.2197$
# Answer:
3. The test - statistic for this data = $1.567$
4. The p - value for this sample = $0.2197$