Question

Homes in a certain town have a mean value of $88,950. It is assumed that homes in the vicinity of the school have a higher value. A sample of 12 homes near the school is selected and it appears as if the population is normal. Their mean value is $92,460 with a standard deviation of $5200. Can we prove with 95% certainty that homes near the school do indeed have a higher value?

Answer 1

Solution :

Null and alternative hypotheses :

The null and alternative hypotheses would be as follows:

Test statistic :

To test the hypothesis the most appropriate test is one sample t-test. The test statistic is given as follows :

Where, x̅ is sample mean, μ is hypothesized value of population mean under H_{​​​​​​0}, s is sample standard deviation and n is sample size.

We have,  x̅ = $92460, μ = $88950, s = $5200 and n = 12

The value of the test statistic is 2.3383.

P-value :

Since, our test is right-tailed test, therefore we shall obtain right-tailed p-value for the test statistic. The right-tailed p-value is given as follows :

P-value = P(T > t)

P-value = P(T > 2.3383)

P-value = 0.0196

The p-value is 0.0196.

Decision :

Level of certainty = confidence level = 95%

Significance level = (100% - 95%) = 5% = 0.05

P-value = 0.0196

(0.0196 < 0.05)

Since, p-value is less than the significance level of 0.05, therefore we shall reject the null hypothesis (H_{​​​​​​0}) at 0.05 significance level.

Conclusion :

At 95% certainty level, there is sufficient evidence to conclude that homes near the school do indeed have a higher value.

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