Question

Ten individuals are chosen at random, from a normal population and their weights (in kg) are
found to be 63, 63, 66, 67, 68, 69, 70, 70, 71 and 71. In the light of this data set, test the claim
that the mean weight in population is 66 kg at 5% level of significance.

Answer 1

Answer on Question #41869 – Math – Statistics and Probability

The sample mean is $\bar{x} = \frac{63 + 63 + 66 + 67 + 68 + 69 + 70 + 70 + 71 + 71}{10} = \frac{678}{10} = 67.8$ (or via an Excel function $= AVERAGE(63; 63; 66; 67; 68; 69; 70; 70; 71; 71)$ )

The sample standard deviation is

$$\begin{array}{l} s = \sqrt {\frac {\sum_ {i = 1} ^ {n} (x _ {i} - \bar {x}) ^ {2}}{n - 1}} = \\ = \sqrt {\frac {2 (6 3 - 6 7 . 8) ^ {2} + (6 6 - 6 7 . 8) ^ {2} + (6 7 - 6 7 . 8) ^ {2} + (6 8 - 6 7 . 8) ^ {2} + (6 9 - 6 7 . 8) ^ {2} + 2 (7 0 - 6 7 . 8) ^ {2} + 2 (7 1 - 6 7 . 8) ^ {2}}{9}} = 3. 0 1 \text {(o r v i a n E x c e l} \\ \end{array}$$

function $= \text{STDEV}(63; 63; 66; 67; 68; 69; 70; 70; 71; 71)$ .

The formulation of the null and alternative hypotheses should be

$H_0: \mu = 66$ versus $H_1: \mu \neq 66$ .

The $t$ test statistic is

$T = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{67.8 - 66}{\frac{3.01}{\sqrt{10}}} = 1.89$ degrees of freedom d.f. $= 9$

Method 1

We test at the level of significance $\alpha = 0.05$ . Since $H_{1}$ is two-tailed, we set the rejection region

$R\colon |T|\geq t_{0.025}$

From the $t$ table we find that $t_{0.025}$ with d.f. $= 9$ is 2.262. Because the observed value $t = 1.89$ is smaller than 2.262, the null hypothesis is not rejected at $\alpha = 0.05$ .

Conclusion: there is strong evidence that the mean weight in population is $66\mathrm{kg}$ (with $\alpha = 0.05$ ).

Method 2

In two-tailed test p-value is the sum of area in two tails, so

$$p - value = P (t \leq - t _ {calculated}) + P (t \geq t _ {calculated}) = P (t \leq - 1.89) + P (t \geq 1.89) =$$

$= 2P(t \geq 1.89) = 0.09 > 0.05 = \alpha$ . To find it, we calculate via excel the function TDIST(1,89;9;2) or seek from t tables with 9 degrees of freedom $t_{0.05} = 1.833$ and $t_{0.025} = 2.262$ , therefore the p-value is between $2^{*}0.05 = 0.1$ and $2^{*}0.025 = 0.05$ , in any case it is greater than 0.05. We do not reject the null hypothesis, because p-value is greater than $\alpha = 0.05$ .

Conclusion: there is strong evidence that the mean weight in population is $66\mathrm{kg}$ (with $\alpha = 0.05$ ).

Method 3

A $95\%$ confidence interval for $\mu$ is

$$\left(\bar {x} - t _ {\frac {\alpha}{2}} \frac {s}{\sqrt {n}}; \bar {x} + t _ {\frac {\alpha}{2}} \frac {s}{\sqrt {n}}\right) = \left(6 7. 8 - 2. 2 6 2 * \frac {3 . 0 1}{\sqrt {1 0}}; 6 7. 8 + 2. 2 6 2 * \frac {3 . 0 1}{\sqrt {1 0}}\right) = (6 5. 6 5; 6 9. 9 5)$$

We can see $\mu_0 = 66$ lies within the $95\%$ confidence interval. The null hypothesis will not be rejected at level $\alpha = 0.05$ if $\mu_0$ lies within the $95\%$ confidence interval.

Conclusion: there is strong evidence that the mean weight in population is $66\mathrm{kg}$ (with $\alpha = 0.05$ ).

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