Saathi Date x.- 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 $5\%$ Significant level assuming that for $9$ degrees of freedom is $1.833$ .

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Q. Saathi Date x.- 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\%

Significant level assuming that for

9

degrees of freedom is

1.833

Calculate Sample 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 males. $\newline$ Sample mean = $\frac{70 + 67 + 62 + 68 + 61 + 68 + 70 + 64 + 64 + 66}{10}$ $\newline$ Sample mean = $\frac{660}{10}$ $\newline$ Sample mean = $66$ inches
Calculate Standard Deviation: Calculate the sample standard deviation. $\newline$ First, find the differences between each height and the sample mean, square these differences, sum them up, and then divide by the number of observations minus one $(n-1)$ to get the variance. Finally, take the square root of the variance to get the standard deviation. $\newline$ Differences squared: $(70-66)^2 + (67-66)^2 + (62-66)^2 + (68-66)^2 + (61-66)^2 + (68-66)^2 + (70-66)^2 + (64-66)^2 + (64-66)^2 + (66-66)^2$ $\newline$ Differences squared: $16 + 1 + 16 + 4 + 25 + 4 + 16 + 4 + 4 + 0$ $\newline$ Sum of differences squared: $90$ $\newline$ Variance: $90 / (10-1)$ $\newline$ Variance: $90 / 9$ $\newline$ Variance: $10$ $\newline$ Standard deviation: $\sqrt{10}$ $\newline$ Standard deviation $\approx 3.16$ inches
Calculate T-Statistic: Calculate the t-statistic to test the hypothesis that the average height is $64$ inches. $\newline$ The t-statistic is calculated using the formula: $t = (\text{sample mean} - \text{hypothesized mean}) / (\text{standard deviation} / \sqrt{n})$ $\newline$ $t = (66 - 64) / (3.16 / \sqrt{10})$ $\newline$ $t = 2 / (3.16 / \sqrt{10})$ $\newline$ $t = 2 / (3.16 / 3.16)$ $\newline$ $t = 2 / 1$ $\newline$ $t = 2$
Compare T-Statistic to Critical Value: Compare the calculated $t$ -statistic to the critical $t$ -value. The critical $t$ -value for a $5\%$ significance level and $9$ degrees of freedom is given as $1.833$ . Since our calculated $t$ -statistic ( $2$ ) is greater than the critical $t$ -value ( $1.833$ ), we reject the null hypothesis that the average height is $t$ $0$ inches.