Question

The heights of 10 males 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 greater than 64 inches Test at 5%.​

Answer 1

Given:

Given that, height of 10 males are found to be 70, 67, 62, 68, 61, 68, 70, 64, 64, 66 inches.

To find:

We need to find the average height is greater than 64 inches.

Solution:

Let us first try to find the idea so that we can confidently reach to a solution.

$x$ = $\frac{\sum x_i}{n}$ = $\frac{660}{10}$ = 66.

∴ $x$ = 66

$x_{i}$                

70                  4                  16

67                   1                   1

62                 -4                  16

68                  2                   4

61                  -5                  25

68                  2                   4

70                  4                   16

64                 -2                   4

64                 -2                   4

66                 0                    0

Now let's write all the given and derived values:

$\sum$ $x_{i}$   = 660

$\sum$ ( $x_{i}$ - $x$ $)^2$ = 90

$S^2 =$ $\frac{\sum (x_i - x)^2}{n-1}$ = $\frac{90}{9}$ = 10

$S = \sqrt{10 } = 3.16$

$\mu$ = 64

Let $H_0$ = The average height is equal to 64 inches

$H_1$ = Average height is more than 64 inches

$t = \frac{x- \mu}{S/\sqrt{n} }$ with the degree n-1 of freedom

t = $\frac{66 - 64}{3.16/ \sqrt{10} }$ = $\frac{2}{1}$

t = 2

The critical value for for a right tailed test at 5% level of significance with degrees of freedom is 1.833.

Calculated value = 2 and Tabulated value = 1.833

If |Calculated value| $\leq$ Tabular value then  $H_0$ is accepted, otherwise rejected.

Now,

But |2| > 1.83

∴ $H_0$ is rejected

So, the average height is greater than 64 inches.

Answer 2

Answer:

answer is

Step-by-step explanation:

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