Question

A survey of a sample of 26 hotels in New Orleans found that the average hotel room rate s $88.42 with a standard deviation of $5.62 and another survey of 25 hotels in the Phoenix area found that the average room rate is $80.61. Assume with a standard deviation of $4.83. At α =  0.05, can it be concluded that there is a significant difference in the rates?
Source: USA TODAY.
Make sure to state Ho, H1 and where the claim is
State the value of the test statistic
Calculate the p-value
Make a decision to reject Ho or fail to reject Ho
State your conclusion in words in the context of the claim
You may use your TI 84 but make sure to state the name of the test you used.
When using calculator and when it comes to choose Yes or No for pooled, keep it at the default No.
Here is what pooled means: If we assume that population variances are equal we select YES and if we assume that population variances are not equal, we select No. By default, use No unless if it is specified otherwise.

Answer 1

Solution:-

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

Null hypothesis: u₁ = u ₂
Alternative hypothesis: u₁ u ₂

Note that these hypotheses constitute a two-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).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 1.4656
DF = 49
t = [ (x₁ - x₂) - d ] / SE

t = 5.33

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 hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 40 degrees of freedom is more extreme than 5.33; that is, less than -5.33 or greater than 5.33.

Thus, the P-value = less than 0.001

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we cannot accept the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that there is a significant difference in the rates.