Question

The Nero Match Company sells matchboxes that are supposed to
have an average of 40 matches per box, with *σ* = 8. A
random sample of 96 matchboxes shows the average number of matches
per box to be 42.3. Using a 1% level of significance, can you say
that the average number of matches per box is more than 40?

What are we testing in this problem?

single meansingle proportion

(a) What is the level of significance?

State the null and alternate hypotheses.

*H*_{0}: *μ* = 40; *H*_{1}:
*μ* < 40*H*_{0}: *μ* = 40;
*H*_{1}: *μ* ≠
40 *H*_{0}: *p* = 40;
*H*_{1}: *p* > 40*H*_{0}:
*p* = 40; *H*_{1}: *p* <
40*H*_{0}: *p* = 40; *H*_{1}:
*p* ≠ 40*H*_{0}: *μ* = 40;
*H*_{1}: *μ* > 40

(b) What sampling distribution will you use? What assumptions are
you making?

The standard normal, since we assume that *x* has a
normal distribution with known *σ*.The Student's *t*,
since we assume that *x* has a normal distribution with
unknown *σ*. The Student's
*t*, since we assume that *x* has a normal
distribution with known *σ*.The standard normal, since we
assume that *x* has a normal distribution with unknown
*σ*.

What is the value of the sample test statistic? (Round your answer
to two decimal places.)

(c) Find (or estimate) the *P*-value.

*P*-value > 0.2500.125 < *P*-value <
0.250 0.050 < *P*-value <
0.1250.025 < *P*-value < 0.0500.005 <
*P*-value < 0.025*P*-value < 0.005

Sketch the sampling distribution and show the area corresponding to
the *P*-value.

(d) Based on your answers in parts (a) to (c), will you reject or
fail to reject the null hypothesis? Are the data statistically
significant at level *α*?

At the *α* = 0.01 level, we reject the null hypothesis
and conclude the data are statistically significant.At the
*α* = 0.01 level, we reject the null hypothesis and conclude
the data are not statistically
significant. At the *α* = 0.01 level,
we fail to reject the null hypothesis and conclude the data are
statistically significant.At the *α* = 0.01 level, we fail
to reject the null hypothesis and conclude the data are not
statistically significant.

(e) Interpret your conclusion in the context of the
application.

There is sufficient evidence at the 0.01 level to conclude that the average number of matches per box is now greater than 40.There is insufficient evidence at the 0.01 level to conclude that the average number of matches per box is now greater than 40.

Answer #1

The Nero Match Company sells matchboxes that are supposed to
have an average of 40 matches per box, with σ = 10. A
random sample of 94 matchboxes shows the average number of matches
per box to be 42.9. Using a 1% level of significance, can you say
that the average number of matches per box is more than 40?
What are we testing in this problem?
single proportionsingle mean
What is the level of significance?
State the null and...

The Nero Match Company sells matchboxes that are supposed to
have an average of 40 matches per box, with σ = 7. A
random sample of 92 matchboxes shows the average number of matches
per box to be 42.7. Using a 1% level of significance, can you say
that the average number of matches per box is more than 40?
What are we testing in this problem?
single mean
single proportion
(a) What is the level of significance?
State the...

The Nero Match Company sells matchboxes that are supposed to
have an average of 40 matches per box, with σ = 8. A
random sample of 96 matchboxes shows the average number of matches
per box to be 42.5. Using a 1% level of significance, can you say
that the average number of matches per box is more than 40?
What are we testing in this problem?
A.) single mean
B.) single
proportion
(a) What is the level of significance?...

The Nero Match Company sells matchboxes that are supposed to
have an average of 40 matches per box, with σ = 11. A
random sample of 95 matchboxes shows the average number of matches
per box to be 43.0. Using a 1% level of significance, can you say
that the average number of matches per box is more than 40?
What are we testing in this problem?
single meansingle proportion
(a) What is the level of significance?
State the null...

The Toylot company makes an electric train with a motor that it
claims will draw an average of only 0.8 ampere (A) under a normal
load. A sample of nine motors was tested, and it was found that the
mean current was x = 1.40 A, with a sample standard
deviation of s = 0.46 A. Do the data indicate that the
Toylot claim of 0.8 A is too low? (Use a 1% level of
significance.)
What are we testing...

The Toylot company makes an electric train with a motor that it
claims will draw an average of no more than 0.8 amps under
a normal load. A sample of nine motors were tested, and it was
found that the mean current was x = 1.30 amps, with a
sample standard deviation of s = 0.43 amps. Perform a
statistical test of the Toylot claim. (Use a 1% level of
significance.)
What are we testing in this problem?
single proportionsingle...

Let x be a random variable representing dividend yield
of bank stocks. We may assume that x has a normal
distribution with σ = 2.8%. A random sample of 10 bank
stocks gave the following yields (in percents).
5.7
4.8
6.0
4.9
4.0
3.4
6.5
7.1
5.3
6.1
The sample mean is x = 5.38%. Suppose that for the
entire stock market, the mean dividend yield is μ = 4.7%.
Do these data indicate that the dividend yield of all...

Let x be a random variable representing dividend yield
of bank stocks. We may assume that x has a normal
distribution with σ = 3.1%. A random sample of 10 bank
stocks gave the following yields (in percents).
5.7
4.8
6.0
4.9
4.0
3.4
6.5
7.1
5.3
6.1
The sample mean is x = 5.38%. Suppose that for the
entire stock market, the mean dividend yield is μ = 5.0%.
Do these data indicate that the dividend yield of all...

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of bank stocks. We may assume that x has a normal
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5.7
4.8
6.0
4.9
4.0
3.4
6.5
7.1
5.3
6.1
The sample mean is x = 5.38%. Suppose that for the
entire stock market, the mean dividend yield is μ = 4.5%.
Do these data indicate that the dividend yield of all...

Let x be a random variable representing
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normal distribution with σ = 2.0%. A random sample of 10 bank
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5.74.86.04.94.03.46.57.15.36.1
The sample mean is x = 5.38%. Suppose that for the entire stock
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