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2 years ago

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A new centrifugal pump is being considered for an application involving the pumping of ammonia. The specification is that the fl

ow rate be more than 5 gallons per minute (gpm). In an initial study, eight runs were made. The average flow rate was 6.5 gpm and the standard deviation was 1.9 gpm. If the mean flow rate is found to meet the specification, the pump will be put into service. What is the P-value

1 answer:

ollegr [7]2 years ago

3 0

Answer:

from the t-distribution table, at df = 7 and t = 2.23

Lies p-values [ 0.05 and 0.025 ]

Hence;

0.025 < p-value < 0.05

Step-by-step explanation:

Given that;

$x^{bar}$ = 6.5 gpm

μ = 5 gpm

n = eight runs = 8

standard deviation σ = 1.9 gpm

Test statistics;

t = ( $x^{bar}$ - μ) / $\frac{s}{\sqrt{n} }$

we substitute

t = (6.5 - 5) / $\frac{1.9}{\sqrt{8} }$

t = 1.5 / 0.67175

t = 2.23

the degree of freedom df = n-1 = 8 - 1

df = 7

Now, from the t-distribution table, at df = 7 and t = 2.23

Lies p-values [ 0.05 and 0.025 ]

Hence;

0.025 < p-value < 0.05

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3 years ago

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(a) 80% confidence interval for the population mean is [109.24 , 116.76].

(b) 80% confidence interval for the population mean is [109.86 , 116.14].

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(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed.

Step-by-step explanation:

We are given that a simple random sample of size n is drawn from a population that is normally distributed.

The sample mean is found to be 113 and the sample standard deviation is found to be 10.

(a) The sample size given is n = 13.

Firstly, the pivotal quantity for 80% confidence interval for the population mean is given by;

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

where, $\bar X$ = sample mean = 113

s = sample standard deviation = 10

n = sample size = 13

$\mu$ = population mean

<em>Here for constructing 80% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.</em>

<u>So, 80% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

P(-1.356 < $t_1_2$ < 1.356) = 0.80 {As the critical value of t at 12 degree

of freedom are -1.356 & 1.356 with P = 10%}

P(-1.356 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.356) = 0.80

P( $-1.356 \times }{\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

<u>80% confidence interval for </u> $\mu$ = [ $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.356 \times }{\frac{10}{\sqrt{13} } }$ , $113+1.356 \times }{\frac{10}{\sqrt{13} } }$ ]

= [109.24 , 116.76]

Therefore, 80% confidence interval for the population mean is [109.24 , 116.76].

(b) Now, the sample size has been changed to 18, i.e; n = 18.

So, the critical values of t at 17 degree of freedom would now be -1.333 & 1.333 with P = 10%.

<u>80% confidence interval for </u> $\mu$ = [ $\bar X-1.333 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.333 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.333 \times }{\frac{10}{\sqrt{18} } }$ , $113+1.333 \times }{\frac{10}{\sqrt{18} } }$ ]

= [109.86 , 116.14]

Therefore, 80% confidence interval for the population mean is [109.86 , 116.14].

(c) Now, we have to construct 98% confidence interval with sample size, n = 13.

So, the critical values of t at 12 degree of freedom would now be -2.681 & 2.681 with P = 1%.

<u>98% confidence interval for </u> $\mu$ = [ $\bar X-2.681 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+2.681 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-2.681 \times }{\frac{10}{\sqrt{13} } }$ , $113+2.681 \times }{\frac{10}{\sqrt{13} } }$ ]

= [105.56 , 120.44]

Therefore, 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed because t test statistics is used only when the data follows normal distribution.

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