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Mekhanik [1.2K]
3 years ago
9

A carpenter is making doors that are 20582058 millimeters tall. If the doors are too long they must be trimmed, and if they are

too short they cannot be used. A sample of 1010 doors is made, and it is found that they have a mean of 20462046 millimeters with a standard deviation of 1515. Is there evidence at the 0.050.05 level that the doors are too short and unusable
Mathematics
1 answer:
melamori03 [73]3 years ago
7 0

Answer:

Z= 0.253

Z∝/2 = ± 1.96

Step-by-step explanation:

Formulate the null and alternative hypotheses as

H0 : u1= u2 against Ha : u1≠ u2 This is a two sided test

Here ∝= 0.005

For alpha by 2 for a two tailed test Z∝/2 = ± 1.96

Standard deviation = s= 15

n= 10

The test statistic used here is

Z = x- x`/ s/√n

Z= 2058- 2046 / 15 / √10

Z= 0.253

Since the calculated value of Z= 0.253 falls in the critical region we reject the null hypothesis.

There is  evidence at the 0.05 level that the doors are too short and unusable.

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a_sh-v [17]

Answer:

Check the explanation

Step-by-step explanation:

Let \overline{x} and \overline{y} be sample means of white and Jesse denotes are two random variables.

Given that both samples are having normally distributed.

Assume \overline{x} having with mean \mu_{1} and \overline{y} having mean \mu_{2}

Also we have given the variance is constant

A)

We can test hypothesis as

H0: \mu_{1} =  \mu_{2}H1: \mu_{1} > \mu_{2}

For this problem

Test statistic is

T=\frac{\overline {x}-\overline {y}}{s\sqrt{\frac {1}{n1} +\frac{1}{n2}}}

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By calculations we get

s=2.41

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Here test statistic is having t-distribution with df=(10+7-2)=15

So p-value is P(t15>2.52)=0.012

Here significance level is 0.05

Since p-value is <0.05 we are rejecting null hypothesis at 95% confidence.

We can conclude that White has significant higher mean than Jesse. This claim we can made at 95% confidence.

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