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Wewaii [24]
3 years ago
10

You believe the population of difference scores is normally distributed, but you do not know the standard deviation. You obtain

pre-test and post-test samples for n = 50 subjects. The average difference (post - pre) is ¯ d = 7.3 with a standard deviation of the differences of s d = 28.7 .What is the test statistic for this sample? (Report answer accurate to three decimal places.)test statistic =______________-What is the p-value for this sample? (Report answer accurate to four decimal places.)
Mathematics
1 answer:
Feliz [49]3 years ago
4 0

Answer:

test stat=1.7986

p value =0.0391

Step-by-step explanation:

given

n=50

d = 7.3

s d = 28.7

formula to calculate test statistic

t=\frac{d}{\frac{s d}{\sqrt{n} } }

⇒ t=\frac{7.3}{\frac{28.7}{\sqrt{50} } }

⇒t=1.7986

for 49 df and t=1.7986  ,p value can be obtained from software

p value=0.0391

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The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 46 o
meriva

Answer:

(a) 68% of the widget weights lie between <u>43 ounces</u> and <u>49 ounces</u>.

(b) The percentage of the widget weights lie between 43 and 87 ounces is 15.87%.

(c) The percentage of the widget weights lie below 76 is 100%.

Step-by-step explanation:

Let <em>X</em> = weight of widgets manufactured by Acme Company.

The distribution of the random variable <em>X</em> is, N (<em>μ </em>= 46, <em>σ</em>²<em> </em>=<em> </em>3²).

According to the Empirical Rule in a normal distribution with mean <em>µ</em> and standard deviation <em>σ</em>, nearly all the data will fall within 3 standard deviations of the mean. The empirical rule can be broken into three parts:

  • 68% data falls within 1 standard deviation of the mean.                       That is P (µ - σ ≤ X ≤ µ + σ) = 0.68.
  • 95% data falls within 2 standard deviations of the mean.                   That is P (µ - 2σ ≤ X ≤ µ + 2σ) = 0.95.
  • 99.7% data falls within 3 standard deviations of the mean.                   That is P (µ - 3σ ≤ X ≤ µ + 3σ) = 0.997.

(a)

According to the Empirical rule, 68% data falls within 1 standard deviation of the mean.

P (µ - σ ≤ X ≤ µ + σ) = 0.68.

Compute the upper and lower values as follows:

<em>µ</em> - <em>σ</em> = 46 - 3 = 43 ounces

<em>µ</em> + <em>σ</em> = 46 + 3 = 49 ounces

Thus, 68% of the widget weights lie between <u>43 ounces</u> and <u>49 ounces</u>.

(b)

Compute the probability of the widget weights lie between 43 and 87 ounces as follows:

P(43

                          =P(-1

*Use a <em>z</em>-table.

The percentage is, 0.1587 × 100 = 15.87%.

Thus, the percentage of the widget weights lie between 43 and 87 ounces is 15.87%.

(c)

Compute the probability of the widget weights lie below 76 as follows:

P(X

                  =P(Z

*Use a <em>z</em>-table.

The percentage is, 1 × 100 = 100%.

Thus, the percentage of the widget weights lie below 76 is 100%.

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