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jasenka [17]
3 years ago
6

Ross and Annie do chores at their home acreage. Ross mows the acreage lawn every 12 days, and Annie bathes the dog every 21 days

. If Ross and Annie do their chores today, how many days will pass before they both do their chores again on the same day?
Mathematics
1 answer:
Readme [11.4K]3 years ago
5 0
84 days will pay before they both do their chores again on the same day.
<span> <span> </span><span><span> Ross mows the lawn <span>      Annie bathes the dog </span>
</span> <span> 12 x 1 = 12                          21 x 1 =  21
</span> <span> 12 x 2 = 24                          21 x 2 =  42
</span> <span> 12 x 3 = 36                          21 x 3 =  63
</span> <span> 12 x 4 = 48                          21 x 4 =  84
</span> <span> 12 x 5 = 60                          21 x 5 = 105
</span> <span> 12 x 6 = 72                          21 x 6 = 126
</span> <span> 12 x 7 = 84                          21 x 7 = 147
</span> <span> 12 x 8 = 96                          21 x 8 = 168
</span> <span> 12 x 9  = 108                       21 x 9 = 189
</span> <span> 12 x 10 = 120                      21 x 10 = 210 </span></span></span>
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A sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.
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(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.

<u><em>Let </em></u>\mu<u><em> = mean of all account balances</em></u>

(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 <u>One-sample t test statistics</u> 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, <u>test statistics</u>  =  \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) <u>Now, P-value of the test statistics is given by the following formula;</u>

          P-value = P( t_8_0 > 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.

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