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AleksAgata [21]
3 years ago
15

A lumber company is making doors that are 2058.0 millimeters tall. If the doors are too long they must be trimmed, and if the do

ors are too short they cannot be used. A sample of 11 is made, and it is found that they have a mean of 2043.0 millimeters with a standard deviation of 25.0. A level of significance of 0.05 will be used to determine if the doors are either too long or too short. Assume the population distribution is approximately normal. Is there sufficient evidence to support the claim that the doors are either too long or too short?
Mathematics
1 answer:
USPshnik [31]3 years ago
7 0

Answer:

No, we don't have sufficient evidence to support the claim that the doors are either too long or too short.

Step-by-step explanation:

We are given that a lumber company is making doors that are 2058.0 millimeters tall. If the doors are too long they must be trimmed, and if the doors are too short they cannot be used.

A sample of 11 is made, and it is found that they have a mean of 2043.0 millimeters with a standard deviation of 25.0.

Let \mu = <u><em>population mean length of doors</em></u>.

So, Null Hypothesis, H_0 : \mu = 2058.0 millimeters      {means that the mean length of doors is 2058.0 millimeters}

Alternate Hypothesis, H_A : \mu\neq 2058.0 millimeters     {means that the doors are either too long or too short}

The test statistics that will be used here is <u>One-sample t-test statistics</u> because we know about the population standard deviation;

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

where, \bar X = sample mean = 2043.0 millimeters

             s = sample standard deviation = 25.0 millimeters

             n = sample of doors = 11

So, <u>the test statistics</u> =  \frac{2043.0-2058.0}{\frac{25.0}{\sqrt{11} } }

                                     =  -1.989  

The value of t-test statistics is -1.989.

Now, at a 5% level of significance, the t table gives a critical value between -2.228 and 2.228 at 10 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so <u><em>we have insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that the mean length of doors is 2058.0 millimeters.

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Step-by-step explanation:

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