Historically, a certain region has experienced 65 thunder days annually. (A "thunder day" is day on which at least one instance of thunder is audible to a normal human ear). Over the past eleven years, the mean number of thunder days is 55 with a standard deviation of 20. Can you conclude that the mean number of thunder days is less than 65? Use the a = 0.01 level of significance.

We are given that Historically, a certain region has experienced 65 thunder days annually. Over the past eleven years, the mean number of thunder days is 55 with a standard deviation of 20.

We have to test that the mean number of thunder days is less than 65 or not.

Let, NULL HYPOTHESIS, [tex]H_0[/tex] : [tex]\mu \geq[/tex] 65 days {means that the mean number of thunder days is higher than or equal to 65}

ALTERNATE HYPOTHESIS, [tex]H_1[/tex] : [tex]\mu[/tex] < 65 days {means that the mean number of thunder days is less than 65}

The test statistics that will be used here is One-sample t-test;

T.S. = [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample mean number of thunder days = 55

s = sample standard deviation = 20

n = number of years = 11

So, test statistics = [tex]\frac{55-65}{\frac{20}{\sqrt{11} } }[/tex] ~ [tex]t_1_0[/tex]

= -1.658

Now, at 1% significance level t table gives critical value of -2.764. Since our test statistics is higher than the critical value of t so we have insufficient evidence to reject null hypothesis as it will not fall in the left side of critical value rejection region.

Therefore, we conclude that the mean number of thunder days is higher than or equal to 65.

shristiparmar1221 shristiparmar1221 · Answer 2 · 2021-12-15T05:29:50+01:00

This question is based on the statistics. Therefore, we have to conclude that the mean number of thunder days is higher than or equal to 65.

Given:

Historically, a certain region has experienced 65 thunder days annually. Over the past eleven years, the mean number of thunder days is 55 with a standard deviation of 20.

We need to determined that the mean number of thunder days is less than 65 or not.

According to question,

Let null hypothesis, [tex]H_0 = \mu\geq 65[/tex] days {the mean number of thunder days is higher than or equal to 65}

Now, alternate hypothesis,[tex]H_0 = \mu<65[/tex] days {means that the mean number of thunder days is less than 65}

Therefore,the test statistics that will be used here is one-sample t-test;

[tex]T.S. = \dfrac{\overline{X}-\mu}{\frac{s}{\sqrt{n} } } \approx t_{n-1}[/tex]

Where, [tex]\overline{X}[/tex] = sample mean number of thunder days = 55

s = sample standard deviation = 20

n = number of years = 11

So, test statistics will be,

[tex]T.S. = \dfrac{55-65}{\frac{20}{\sqrt{11} } } \approx t_{10}[/tex]

= -1.658

Thus, at 1% significance level t table gives critical value of -2.764.

So, our test statistics is higher than the critical value of t . Thus, we have insufficient evidence to reject null hypothesis as it will not fall in the left side of critical value rejection region.

Therefore, we have to conclude that the mean number of thunder days is higher than or equal to 65.

For more details, prefer this link:

https://brainly.com/question/20165896