Evaluate the expression for x=1 and y=9 3x+y=
2 answers:
You might be interested in
(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.
know more about test statics visit this link
#SPJ4
1. Assume the manufacturer produced 100 products.
2. 70 are acceptable and 30 are defective.
3. the first product has a probability of 70/100 to be selected from the acceptable ones. The second 69/99 (one is removed from the good ones) and the third one 68/98
4. so P(pick 3 consecutive acceptable products)=
=0.339
5. If familiar to the Combination formula C(n, r)=
:
P(picking 3 acceptable out of 100)=n(picking 3 acceptable)/n(picking 3)=C(70, 3)/C(100/3)=
= 0.339
2. 70 are acceptable and 30 are defective.
3. the first product has a probability of 70/100 to be selected from the acceptable ones. The second 69/99 (one is removed from the good ones) and the third one 68/98
4. so P(pick 3 consecutive acceptable products)=
5. If familiar to the Combination formula C(n, r)=
P(picking 3 acceptable out of 100)=n(picking 3 acceptable)/n(picking 3)=C(70, 3)/C(100/3)=