Chapter 8.2, Problem 22P

(i)

To determine

Check the sample mean by using the calculator.

Check the sample standard deviation by using the calculator.

Explanation of Solution

Calculation:

Let x denotes tree-ring dates in this excavation area.

Mean:

Use Ti 83 calculator to find the mean as follows:

• Select STAT > Edit > Enter the values of Dates.
• Click $\boxed{2^{\text{nd}}}$ button; take the arrow to the MATH menu, and then ‘3’ numbered key.
• Click $\boxed{2^{\text{nd}}}$ button; then ‘1’ numbered key to get L1, and close the ‘)’ bracket.
• Click Enter.

Output using Ti 83 calculator is given below:

From the Ti 83 calculator output, the mean value is 1268.

Standard deviation:

Use Ti 83 calculator to find the standard deviation as follows:

• Select STAT > Edit > Enter the values of Dates.
• Click $\boxed{2^{\text{nd}}}$ button; take the arrow to the MATH menu, and then ‘7’ numbered key.
• Click $\boxed{2^{\text{nd}}}$ button; then ‘1’ numbered key to get L1, and close the ‘)’ bracket.
• Click Enter.

Output using Ti 83 calculator is given below:

From the Ti 83 calculator output, the standard deviation value is approximately 37.29.

Hence, the sample mean and the sample standard deviation by using the calculator is verified.

(ii)

To determine

(a)

Find the level of significance.

State the null and alternative hypothesis.

(b)

Identify the sampling distribution to be used.

Explain how the sampling distribution is chosen.

Find the sampling distribution value of the sample test statistic.

(c)

Find the P-value.

Draw the sampling distribution by showing the area corresponding to the P-value.

(d)

Check whether the null hypothesis is rejecting or fail to reject.

Identify whether the data statistically significant at level 0.01 or not.

(e)

Interpret the conclusion in the context of the application.

Answer to Problem 22P

(a)

The level of significance is 0.01.

The null hypothesis is $H_0:\mu =1300$.

The alternative hypothesis is $H_1:\mu \ne 1300$.

(b)

The sampling distribution to be used is Student’s t distribution.

The Student’s t distribution is chosen because the x distribution is normal and population standard deviation is not known.

The sampling distribution value of the sample test statistic is –2.714.

(c)

The P-value is 0.0239.

(d)

The null hypothesis is failed to be rejected.

The data is not statistically significant at level 0.01.

Explanation of Solution

Calculation:

(a)

Let $\mu$ denotes the mean of tree-ring dates in the area.

From the given information the value of $\alpha$ is 0.01, and the population mean of tree-ring dates in the area is different from that in 1300 A.D.

Hence, the level of significance is 0.01.

The null and alternative hypothesis is,

Null hypothesis:

$H_0:\mu =1300$

Alternative hypothesis:

$H_1:\mu \ne 1300$

(b)

Conditions:

• When the x distribution considered in the study has the normal distribution with the known population standard deviation $\sigma$ then the sampling distribution $\overline{x}$ has normal distribution for any sample size n. The standardized z statistic is used for testing.
• When the x distribution considered in the study is not normally distributed and the population standard deviation $\sigma$ is known then the sampling distribution $\overline{x}$ has normal distribution if the sample size n is greater than or equal 30. That is, $n\ge 30$.

Test statistic for t:

The z statistic value for sample test statistic $\overline{x}$ is,

$t=\frac{\overline{x}-\mu }{\left(\frac{s}{\sqrt{n}}\right)}$

In the formula $\overline{x}$ is sample mean, $\mu$ is population mean, $\sigma$ is sample standard deviation, and n is the sample size.

The distribution of x is assumed to be normal and the population standard deviation is unknown. The x distribution considered in the study is normally distributed and population standard deviation is unknown. Hence, the sampling distribution to be used is Student’s t distribution.

T-statistic:

Substitute $\overline{x}$ as 1268, $\mu$ as 1300, s as 37.29, and n as 10 in the test statistic formula

$\begin{array}{c}t=\frac{1268-1300}{\left(\frac{37.29}{\sqrt{10}}\right)}\\ =\frac{-32}{11.7921}\\ =-2.714\end{array}$

Hence, the sampling distribution value of the sample test statistic is –2.714.

(c)

Use Ti 83 calculator to obtain the P-value as follows:

• Select STAT, take the arrow to the TESTS menu, and then ‘2’ numbered key.
• Select Stats under Inpt.
• Enter ${\mu }_0$ as 1300.
• Enter $\overline{x}$ as 1268, $S_x$ as 37.29, and n as 10.
• Select the option $\ne {\mu }_0$ under $\mu$.
• Click Draw.

Output using Ti 83 calculator is given below:

From the Ti 83 calculator output, the P-value is 0.0239.

Hence, the P-value is 0.0239.

(d)

Rejection rule:

• If the P-value is less than or equal to $\alpha$, then reject the null hypothesis and the test is statistically significant. That is, $P\text{-value}\le \alpha$.

Conclusion:

The P-value is 0.0239 and the level of significance is 0.01.

The P-value is greater than the level of significance.

That is, $0.0239\left(=P\text{-value}\right)>0.01\left(=\alpha \right)$.

By the rejection rule, the null hypothesis is failed to be rejected.

Hence, the data is not statistically significant at level 0.01.

(e)

The null hypothesis is failed to be rejected. This shows that, the population mean of tree-ring dates in the area is not different from that in 1300 A.D at level of significance 0.05.