Question

? The height of 10 males of a
given locality are found to be 70, 67,
62, 68, 61, 68, 70, 69, 64, 66 inches.
Assume that average height is 64
inches, then the value of t at 5%
level of significance and 9 degree o
freedom is :​

Answer 1

Answer:

At 0.05 level or 5% level of significance the tabulated value ... The height of 10 males of a given locality is found to be 70, 67, 62, 68, 61, 68, ... i.e., the average height is equal to 64 inches.

Answer 2

Answer:

Concept :

To determine whether the outcomes of an experiment or survey are meaningful, you can test your hypotheses in statistics. Calculating the likelihood that your results were the result of chance is essentially a test of the validity of your findings. The experiment won't be repeatable and won't of much help if your findings may have happened by accident. One of the most perplexing parts of the curriculum for students is hypothesis testing, largely because you have to know your null hypothesis before you can even start the test. It can be challenging to understand those complex word problems you encounter frequently.

Explanation:

Given :

$x$    :         70 67 62 68 61 68 70 64 64 66

$x_{i}$ - $\bar{x}$ :       4    1   -4  2   -5   2  4   -2  -2  0

$(x_{i} -\bar{x}) ^{2}$:  16   1    16  4  25  4  16   4   4  0

∑$x$ = 660

∑$(x_{i} -\bar{x}) ^{2}$ = 90

Solution :

Step 1:

$\bar{x}$ = $\frac{\sum x_{i}}{n}$ =  $\frac{660}{10}$ = 66

$s^{2}$ =  $\frac{\sum(x_{i} -\bar{x}) ^{2} }{n - 1}$ =  $\frac{90}{9}$ = 10 |\lambda|

∴ s =$\sqrt{10}$

     = 2.162

Step 2:

Null hypothesis : $H_{0}$ : $\mu = 64$

Alternative hypothesis :  $H_{1}$ :  $\mu < 64$    

$\lambda$ = $\frac{\bar{x}- \mu }{\frac{s}{\sqrt{n}}}$    

  = $\frac{66- 64}{\frac{3.162}{\sqrt{10}}}$

$|\lambda|$ = 2

Step 3:

• The critical value for 1 to 5

• The level of significance with 10 - 1 = 9

• Degree of freedom is 1.833

Calculated value = 2

Tabulated value = 1.833

But | 2 | > 1.833

∴  Hence we reject $H_{0}$

#SPJ2