A school library has 200 DVDs in its collection. Given that 40% of the DVDs are about science and 45% are about history, how man
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The poisson distribution applies here.
(A.)test statistic for this test using Ha: p8くPN is Z = -2.164 (B.) p-value (3 decimal places) is 0.015. (C.) Based on this p-value, which of the following would you expect for a 95% is negative. (D.) the correct conclusion is The proportion of Boston drivers wearing seat was significantly lower than the proportion of new York drivers wearing seat belts.
A.)
According to given data find the test statistic for this test using Ha:
p1 = 0.581
p2 = 0.832
standard error = SE = 0.116
<h3>Test statistics :-</h3>Z =
Z = = -2.164
Z = -2.164
B.)
According to given Determine the p-value (3 decimal places).
<h3>p - value</h3>it is left tailed test.
p - value for left tailed test is,
p - value = P(Z < test statistics )
p - value = P(Z < -2.164)
P - value = 0.015
C.)
Based on this p-value, which of the following would you expect for a 95%
confidence level = 95% = 0.95
significance level = α = 0.05
P -value is less than 0.05
so, reject null hypothesis at α = 0.05
That is <
so, based on P - value , All values in the 95% confidence interval
PB - PNY is negative.
D.)
According to given data What is the correct conclusion?
Reject null hypothesis at α = 0.05
so, <
Hence, The proportion of Boston drivers wearing seat was significantly lower than the proportion of new York drivers wearing seat belts.
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0.017559
It is given that:
Probability of Households Having cable TV, p₀ = 76% = 0.76
Therefore,
The probability that the Households not having cable TV = 1 - 0.76 = 0.24
Sample size, n = 225 households
sample proportions is less than 82% i.e p = 0.82
Now,
The standard error, SE =
= 0.02847
Z =
Z = 2.107
therefore, P(sample porportions < 0.82) = P(Z < 2.107)
now from the p value from the Z table we get P(sample porportions < 0.82) = 0.017559
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