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Leno4ka [110]
2 years ago
5

Suppose that we want to test the hypothesis with a significance level (alpha) of .05 that the climate has changed since industri

alization. Suppose that the mean temperature throughout history is 50 degrees. During the last 40 years, the mean temperature has been 51 degrees and suppose the population standard deviation (z distributed) is 2 degrees. What can we conclude? Then construct a 95 percent confidence interval for the population mean.
Mathematics
1 answer:
saw5 [17]2 years ago
7 0

Answer: we reject the initial claim

(50. 3802, 51.6198)

Step-by-step explanation:

A)

The initial statement is that the mean temperature throughout history is 50°

This is the null hypothesis and it is denoted as H'

H': u = 50.

After taking a sample of 40 years, we realized that the mean temperature is 51°.

This is the alternative hypothesis, in contradiction to the alternative.

The alternative hypothesis is denoted as H1

H1: u > 50 ( upper tailed).

Sample size (n) = 40

Sample mean (x) = 51

Population standard deviation (σ) = 2

We use a z test to get the value of the test statistics.

We are using a z test because sample size is greater than 30 ( n = 40) and population standard deviation is given.

Z score = x - u/ (σ/√n)

Z score = 51 - 50 / (2/√40)

Z score = 1 / 0.3162

Z score = 3.16

Our level of significance is 5%, and the critical value at this level of significance is 1.645.

By comparing the z score relative to the critical value, since our z score is greater than 1.645, it implies that we are in the rejection region hence we reject the initial claim.

B)

We are to construct a 95% confidence interval for population mean temperature.

For upper limit

u = x + Zα/2 ×(σ/√n)

Where Zα/2 = 1.96 which is the critical value for a two tailed test at 5% level of significance.

For upper limit, we have that

u = 51 + 1.96 × (2/√40)

u = 51 + 1.96 (0.3162)

u = 51 + 0.6198

u = 51.6198.

For lower limit, we have that

u = 51 - 1.96 × (2/√40)

u = 51 - 1.96 (0.3162)

u = 51 - 0.6198

u = 50. 3802.

Hence the 95% confidence level for mean temperature is given as (50. 3802°, 51.6198°)

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