Question

It sampled 40 customers in San Francisco and 50 customers in San Diego to assess potential demand.

On a scale of 1-7 (7 = very likely to buy), San Diego customers had a mean of 3.5 with a standard deviation of 1.1. SF customers had a mean of 4.1 with a standard deviation of 2.3.

Are these markets statistically different?

1.Compute standard error

2.Compute t-calc  

3.Compare |t-calc| to 1.96 (95% confidence in our results) and 2.58 (99% confidence)

4.If |t-calc| > 1.96, reject the null with 95% confidence

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Standard error for 2 means (sxs_x ̅ ) = ?12?1+?22?2√((s_1^2)/n_1 +(s_2^2)/n_2 )

T-calc for 2 means = ?1−?2??(x ̅_1-x ̅_2)/s_x ̅

Answer 1

1 denotes the San Francisco city and 2 denotes the San Diego city

The sample means are shown below:

Also, the sample standard deviations are:

and the sample sizes are n1​=40 and n2​=50.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ = μ2​

Ha: μ1​ ≠ μ2​

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Testing for Equality of Variances

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

The critical values are , and since F = 4.372, then the null hypothesis of equal variances is rejected.

(2) Rejection Region

The significance level is α=0.05, and the degrees of freedom are df=53.162. In fact, the degrees of freedom are computed as follows, assuming that the population variances are unequal:

Hence, it is found that the critical value for this two-tailed test is , for α=0.05 and df = 53.162.

The rejection region for this two-tailed test is R = { t : ∣t∣ > 2.006}.

(3) Test Statistics

Since it is assumed that the population variances are unequal, the t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that ∣t∣ = 1.517 ≤ tc ​= 2.006, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.1352, and since p = 0.1352 ≥ 0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​ is different than μ2​, at the 0.05 significance level.

We can conclude that these markets of San Francisco and San Diego are not statistically different.

Confidence Interval

The 95% confidence interval is −0.193<μ1​−μ2​<1.393.