Question

Edit question Suppose in an effort to manage his inventory levels better, the owner of two steak and seafood restaurants, both located in the same city, hires a statistician to conduct a statistical study. The owner is interested in whether the restaurant located on the south side sells more halibut fillets per night than the restaurant located on the north side of the city. The statistician selects a random sample of 102 nights that the south-side restaurant is open. The mean number of halibut fillets sold per night at the south-side location is 15.3 with a sample standard deviation of 0.4. Likewise, the mean number of halibut fillets sold per night at the random sample of 83 nights that the north-side restaurant is open is 16.8 with a sample standard deviation of 5.5.

Suppose you intend to conduct a hypothesis test on the difference in population means. In preparation, you identify the sample of nights at the south restaurant as sample 1 and the sample of nights at the north restaurant as sample 2. Organize the provided data by completing the following table:

Sample 1 Sample 2 N1 = N2 = μ1 = μ2 = X‾‾1 = X‾‾2 = σ1 = σ2 = s1 = s2 = The difference in sample means for sample 1 and sample 2 is . The estimate of the standard deviation of the sampling distribution of the differences in sample means, s(X‾‾−X‾‾) , is . Now, you know all that you need to know to answer the question about whether the restaurant located on the south side sells more halibut fillets per night than the restaurant located on the north side of the city.

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_South< u_North
Alternative hypothesis: u_South > u_North

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

N₁ = 102

N₂ = 83

X‾‾1 = 15.3

X‾‾2 = 16.8

s1 = 0.40

s2 = 5.5

The difference in the sample means is - 1.50.

The estimate of the standard deviation of the sampling distribution of the differences in sample means is 0.6050.

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 0.6050
DF = 183

t = [ (x₁ - x₂) - d ] / SE

t = - 2.48

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesised difference between population means, and SE is the standard error.

The observed difference in sample means produced a t statistic of - 2.48 .

Therefore, the P-value in this analysis is 0.993 .

Interpret results. Since the P-value (0.993) is greater than the significance level (0.05), we cannot reject the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that the restaurant located on the south side sells more halibut fillets per night than the restaurant located on the north side of the city.