Question

A sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.

Answer 1

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ = $1,150  

    Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ $1,150

(b) The test statistic is 3.571.

(c) We conclude that the mean of all account balances is significantly different from $1,150.

Step-by-step explanation:

We are given that a sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.

We have test the hypothesis to determine whether the mean of all account balances is significantly different from $1,150.

Let $\mu$ = mean of all account balances

(a)So, Null Hypothesis, $H_0$ : $\mu$ = $1,150     {means that the mean of all account balances is equal to $1,150}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ $1,150    {means that the mean of all account balances is significantly different from $1,150}

The test statistics that will be used here is One-sample t test statistics as we don't know about population standard deviation;

                               T.S.  = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$  ~ $t_n_-_1$

where, $\bar X$ = sample average balance = $1,200

             s = sample standard deviation = $126

             n = sample of account balances = 81

(b) So, test statistics  =  $\frac{1,200-1,150}{\frac{126}{\sqrt{81} } }$  ~ $t_8_0$

                               =  3.571

The value of the sample test statistics is 3.571.

(c) Now, P-value of the test statistics is given by the following formula;

          P-value = P( $t_8_0$ > 3.571) = Less than 0.05%

Because in the t table the highest critical value for t at 80 degree of freedom is given between 3.460 and 3.373 at 0.05% level.

Now, since P-value is less than the level of significance as 5% > 0.05%, so we sufficient evidence to reject our null hypothesis.

Therefore, we conclude that the mean of all account balances is significantly different from $1,150.