Question 133760: Hi, Can someone help me with these 2 questions, I'm a little confused as to what I am suppose to be doing.
1. You are the incoming inspector for potato chips you are to ensure that each bag has 16 ounces or more in it. You want your testing to be at the level of significance of 0.05. You pull a sample of 49 bags of chips from a recent truckload. Your sample statistics are:
x-bar (the sample mean) = 15.9 ounces
s ( the sample standard deviation) = 0.35 ounces
(a) what is the null and alternative hypothesis
(b) one or two tailed test ??
(c) what is the critical z value for your test at the 0.05 level of significance??
(d) what is the calculated z value ??
(e) what is your decision about the load of potato chips ??
-- reject ?? -- not-reject ??
(a) Ho : μ <= 16 Vs H1 : μ < 16 (Left tailed test)
(b) One tailed test
(c) Critical z = 1.645
(d) Test Statistics:
z = follows N(0,1)
= -2
where
x bar=15.9 n=49 s=0.35 μ= 16
(e) Rejection Rule
Critical Value = -1.645
Thus we reject H0 if z < -1.645
As z = -2 < -1.645 we reject H0.
At the 5% level of significance, the data provides enough evidence to reject the null hypothesis. Thus we conclude that each bag has less than 16 ounces.
2. An owner of a fast food restaurant reported to corporate headquarters that the average bill paid by his customers in the last quarter was $6.20 and that the standard deviation was $1.90. Not knowing exactly what the effect would be, headquarters suddenly launched a nationwide promotional campaign featuring a large quantity discount for a multi-sandwich purchase. The stubs from the next 81 purchases at the owners franchise after the campaign was launched averaged $6.65.
Conduct the 5 step hypothesis test at a level of significance of 0.05 to determine if the promotion increased the average bill amount
Ans.
To Test
Ho : μ= 6.20 Vs H1 : μ > 6.20 (Right tailed test)
Level of significance = 0.05
Test Statistics:
z = xbar-μ/SE follows N(0,1)
= 2.13
where
xbar =6.65 n=81 sd=1.9 μ= 6.2
P-value = P(z > 2.13) = 0.0165
Since P-value of 0.0165 < 0.05 we reject H0.
It is statistically significant
Rejection Rule
Critical Value = 1.645
Thus we reject H0 if z > 1.645
As z = 2.13 > 1.645 we reject H0.
Conclusion
At the 5% level of significance, the data provides enough evidence to reject the null hypothesis. Thus we conclude that the promotion increased the average bill amount.
Answer by nanaktutors@yahoo.com(8)   (Show Source):
You can put this solution on YOUR website! 1. You are the incoming inspector for potato chips you are to ensure that each bag has 16 ounces or more in it. You want your testing to be at the level of significance of 0.05. You pull a sample of 49 bags of chips from a recent truckload. Your sample statistics are:
x-bar (the sample mean) = 15.9 ounces
s ( the sample standard deviation) = 0.35 ounces
(a) what is the null and alternative hypothesis
(b) one or two tailed test ??
(c) what is the critical z value for your test at the 0.05 level of significance??
(d) what is the calculated z value ??
(e) what is your decision about the load of potato chips ??
-- reject ?? -- not-reject ??
(a) Ho : μ <= 16 Vs H1 : μ < 16 (Left tailed test)
(b) One tailed test
(c) Critical z = 1.645
(d) Test Statistics:
z = follows N(0,1)
= -2
where
x bar=15.9 n=49 s=0.35 μ= 16
(e) Rejection Rule
Critical Value = -1.645
Thus we reject H0 if z < -1.645
As z = -2 < -1.645 we reject H0.
At the 5% level of significance, the data provides enough evidence to reject the null hypothesis. Thus we conclude that each bag has less than 16 ounces.
2. An owner of a fast food restaurant reported to corporate headquarters that the average bill paid by his customers in the last quarter was $6.20 and that the standard deviation was $1.90. Not knowing exactly what the effect would be, headquarters suddenly launched a nationwide promotional campaign featuring a large quantity discount for a multi-sandwich purchase. The stubs from the next 81 purchases at the owners franchise after the campaign was launched averaged $6.65.
Conduct the 5 step hypothesis test at a level of significance of 0.05 to determine if the promotion increased the average bill amount
Ans.
To Test
Ho : μ= 6.20 Vs H1 : μ > 6.20 (Right tailed test)
Level of significance = 0.05
Test Statistics:
z = xbar-μ/SE follows N(0,1)
= 2.13
where
xbar =6.65 n=81 sd=1.9 μ= 6.2
P-value = P(z > 2.13) = 0.0165
Since P-value of 0.0165 < 0.05 we reject H0.
It is statistically significant
Rejection Rule
Critical Value = 1.645
Thus we reject H0 if z > 1.645
As z = 2.13 > 1.645 we reject H0.
Conclusion
At the 5% level of significance, the data provides enough evidence to reject the null hypothesis. Thus we conclude that the promotion increased the average bill amount.
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