```Question 133760
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 ??
-- reject ??     -- not-reject ??
(a) Ho : &#956; <= 16   Vs    H1 : &#956; < 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    &#956;= 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 owners 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 : &#956;= 6.20  Vs    H1 : &#956; > 6.20 (Right tailed test)
	Level of significance = 0.05
	Test Statistics:
z = xbar-&#956;/SE   follows N(0,1)
= 2.13
where
xbar =6.65    n=81    sd=1.9    &#956;= 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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