Question

5. 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% significance level
assuming that for 9 degrees of freedom t = 1.833 at a = 0.05.​

Answer 1

Answer:

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.

xx = \frac{\sum x_i}{n}

n

∑x

i

= \frac{660}{10}

10

660

= 66.

∴ xx = 66

x_{i}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}x

i

= 660

\sum∑ ( x_{i}x

i

- xx )^2)

2

= 90

S^2 =S

2

= \frac{\sum (x_i - x)^2}{n-1}

n−1

∑(x

i

−x)

2

= \frac{90}{9}

9

90

= 10

S = \sqrt{10 } = 3.16S=

10

=3.16

\muμ = 64

Let H_0H

0

= The average height is equal to 64 inches

H_1H

1

= Average height is more than 64 inches

t = \frac{x- \mu}{S/\sqrt{n} }t=

S/

n

x−μ

with the degree n-1 of freedom

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

3.16/

10

66−64

= \frac{2}{1}

1

2

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_0H

0

is accepted, otherwise rejected.

Now,

But |2| > 1.83

∴ H_0H

0

is rejected

So, the average height is greater than 64 inches.

Step-by-step explanation:

i hope its better