Date $\newline$ Saathi The heights of ten mails of a given locality are found to be $70,67,62, 68,61,68,70,64,64,66$ (inches) Is it reasonable to believe that the average height is $64$ inches. Test at $5\%$ Significent level assuming that for $9$ degrees of freedom is $1.833$ ,

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Q. Date

\newline

Saathi The heights of ten mails of a given locality are found to be

70,67,62, 68,61,68,70,64,64,66

(inches) Is it reasonable to believe that the average height is

64

inches. Test at

5\%

Significent level assuming that for

9

degrees of freedom is

1.833

Calculate Mean: Calculate the sample mean (average height) of the ten males. $\newline$ To find the sample mean, add all the heights together and divide by the number of heights. $\newline$ Sample mean = $\frac{70 + 67 + 62 + 68 + 61 + 68 + 70 + 64 + 64 + 66}{10}$ $\newline$ Sample mean = $\frac{650}{10}$ $\newline$ Sample mean = $65$ inches
Set Hypotheses: Set up the null hypothesis and the alternative hypothesis. $\newline$ The null hypothesis ( $H_0$ ) is that the true population mean is $64$ inches. $\newline$ The alternative hypothesis ( $H_1$ ) is that the true population mean is not $64$ inches.
Calculate Standard Deviation: Calculate the standard deviation of the sample. $\newline$ To calculate the standard deviation, use the formula: $\newline$ $s = \sqrt{\frac{\Sigma(xi - \bar{x})^2}{n - 1}}$ $\newline$ where $xi$ is each individual height, $\bar{x}$ is the sample mean, and $n$ is the number of observations. $\newline$ $s = \sqrt{\frac{((70-65)^2 + (67-65)^2 + (62-65)^2 + (68-65)^2 + (61-65)^2 + (68-65)^2 + (70-65)^2 + (64-65)^2 + (64-65)^2 + (66-65)^2)}{9}}$ $\newline$ $s = \sqrt{\frac{(25 + 4 + 9 + 9 + 16 + 9 + 25 + 1 + 1 + 1)}{9}}$ $\newline$ $s = \sqrt{\frac{100}{9}}$ $\newline$ $s = \sqrt{11.111}$ $\newline$ $s \approx 3.33$ inches
Calculate T-Statistic: Calculate the t-statistic using the sample mean, the hypothesized mean, the standard deviation, and the number of observations. $\newline$ The t-statistic is calculated using the formula: $\newline$ $t = \frac{\bar{x} - \mu}{s / \sqrt{n}}$ $\newline$ where $\bar{x}$ is the sample mean, $\mu$ is the hypothesized mean, $s$ is the standard deviation, and $n$ is the number of observations. $\newline$ $t = \frac{65 - 64}{3.33 / \sqrt{10}}$ $\newline$ $t = \frac{1}{3.33 / 3.16}$ $\newline$ $t = \frac{1}{1.053}$ $\newline$ $t \approx 0.95$
Compare T-Statistic: Compare the calculated $t$ -statistic to the critical $t$ -value at the $5\%$ significance level for $9$ degrees of freedom. $\newline$ The critical $t$ -value given is $1.833$ . $\newline$ Since the calculated $t$ -statistic ( $0.95$ ) is less than the critical $t$ -value ( $1.833$ ), we do not reject the null hypothesis.
Make Conclusion: Make a conclusion based on the comparison of the t-statistic and the critical t-value. $\newline$ Since the calculated $t$ -statistic is less than the critical $t$ -value, there is not enough evidence to reject the null hypothesis. Therefore, it is reasonable to believe that the average height could be $64$ inches.