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Lilit [14]
4 years ago
7

Sunspots have been observed for many centuries. Records of sunspots from ancient Persian and Chinese astronomers go back thousan

ds of years. Some archaeologists think sunspot activity may somehow be related to prolonged periods of drought in the southwestern United States. Let x be a random variable representing the average number of sunspots observed in a 4-week period. In a random sample of 40 such periods from Spanish colonial times, the sample mean is x¯ = 47.0. Previous studies of sunspot activity during this period indicate that σ = 35. It is thought that for thousands of years, the mean number of sunspots per 4-week period was about µ = 41. Sunspot activity above this level may (or may not) be linked to gradual climate change. Do the data indicate that the mean sunspot activity during the Spanish colonial period was higher than 41? Use α = 0.05. (a) State the appropriate null and alternate hypothesis. (b) Compute the value of test statistic. (c) Using P-value method, state a conclusion.
Mathematics
1 answer:
Lunna [17]4 years ago
8 0

Answer:

(a) Null Hypothesis, H_0 : \mu \leq 41  

    Alternate Hypothesis, H_A : \mu > 41

(b) The value of z test statistics is 1.08.

(c) We conclude that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41.

Step-by-step explanation:

We are given that in a random sample of 40 such periods from Spanish colonial times, the sample mean is x¯ = 47.0. Previous studies of sunspot activity during this period indicate that σ = 35.

It is thought that for thousands of years, the mean number of sunspots per 4-week period was about µ = 41.

Let \mu = <u><em>mean sunspot activity during the Spanish colonial period.</em></u>

(a) Null Hypothesis, H_0 : \mu \leq 41     {means that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41}

Alternate Hypothesis, H_A : \mu > 41     {means that the mean sunspot activity during the Spanish colonial period was higher than 41}

The test statistics that would be used here <u>One-sample z test statistics</u> as we know about the population standard deviation;

                            T.S. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean = 47

            σ = population standard deviation = 35

            n = sample of periods from Spanish colonial times = 40

So, <em><u>the test statistics</u></em>  =  \frac{47-41}{\frac{35}{\sqrt{40} } }

                                      =  1.08

(b) The value of z test statistics is 1.08.

(c) <u>Now, the P-value of the test statistics is given by;</u>

                P-value = P(Z > 1.08) = 1 - P(Z < 1.08)

                              = 1 - 0.8599 = <u>0.1401</u>

Since, the P-value of the test statistics is higher than the level of significance as 0.1401 > 0.05, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41.

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Answer:

(a) Probability that a single randomly selected element X of the population is between 57,000 and 58,000 = 0.46411

(b) Probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = 0.99621

Step-by-step explanation:

We are given that a normally distributed population has mean 57,800 and standard deviation 75, i.e.; \mu = 57,800  and  \sigma = 750.

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The z probability is given by;

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Now, probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = P(57,000 < X bar < 58,000)

P(57,000 <= X bar <= 58,000) = P(X bar <= 58,000) - P(X bar < 57,000)

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P(X < 57000) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{57000-57800}{\frac{750}{\sqrt{100} } } ) = P(Z < -10.67) = P(Z > 10.67)

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Therefore, P(57,000 < X bar < 58,000) = 0.99621 - 0 = 0.99621 .

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