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Answer:
(a) Null Hypothesis, :
= $1,150
Alternate Hypothesis, :
$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.
<u><em>Let </em></u><u><em> = mean of all account balances</em></u>
(a)So, Null Hypothesis, :
= $1,150 {means that the mean of all account balances is equal to $1,150}
Alternate Hypothesis, :
$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 <u>One-sample t test statistics</u> as we don't know about population standard deviation;
T.S. = ~
where, = sample average balance = $1,200
s = sample standard deviation = $126
n = sample of account balances = 81
(b) So, <u>test statistics</u> = ~
= 3.571
The value of the sample test statistics is 3.571.
(c) <u>Now, P-value of the test statistics is given by the following formula;</u>
P-value = P( > 3.571) = <u>Less than 0.05%</u>
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.
It is given that area of rectangle of Kaitlin is 1500 sq.ft. with length 50 ft and a width of 30 ft.
The new area when new length 2 times the current length and the new width 2 times the current width.
a. The size of Kaitlin new patio is
b. The area of new patio is 4 times the area of current patio.