A metallurgist has one alloy containing 34% copper another containing 48% copper. How many pounds of each alloy must he use to m
1 answer:
Answer:
<h2>36.14 pounds of 34% copper alloy and 9.86 pounds of 48% copper alloy</h2>Step-by-step explanation:
First alloy contains 34% copper and the second alloy contains 48% alloy.
We wish to make 46 pounds of a third alloy containing 37% copper.
Let the weight of first alloy used be in pounds and the weight of second alloy used be
in pounds.
Total weight =
Total weight of copper =
Subtracting 34 times first equation from second equation,
∴ 36.14 pounds of first alloy and 9.86 pounds of second alloy were used.
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Answer:
(a) Null Hypothesis, :
= 22 ounces
Alternate Hypothesis, :
22 ounces
(b) The value of the test statistic is -2.687.
(c) The critical values are -1.96 and 1.96.
(d) We conclude that Reject since the value of the test statistic is less than the negative critical value.
Step-by-step explanation:
We are given that a local bottler in Hawaii wishes to ensure that an average of 22 ounces of passion fruit juice is used to fill each bottle.
He takes a random sample of 65 bottles. The mean weight of the passion fruit juice in the sample is 21.54 ounces. Assume that the population standard deviation is 1.38 ounce.
<em>Let </em><em> = average ounces of passion fruit juice used to fill each bottle.</em>
(a) So, Null Hypothesis, :
= 22 ounces {means that the average of 22 ounces of passion fruit juice is used to fill each bottle}
Alternate Hypothesis, :
22 ounces {means that the average different from 22 ounces of passion fruit juice is used to fill each bottle}
The test statistics that would be used here <u>One-sample z test statistics</u> as we know about the population standard deviation;
T.S. = ~ N(0,1)
where, = sample mean weight of the passion fruit juice = 21.54 ounces
= population standard deviation = 1.38 ounce
n = sample of bottles = 65
So, <u><em>test statistics</em></u> =
= -2.687
(b) The value of z test statistics is -2.687.
(c) Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.
<em>Since our test statistics does not lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><u><em>we reject our null hypothesis</em></u><em>.</em>
Therefore, we conclude that Reject since the value of the test statistic is less than the negative critical value which means that the average different from 22 ounces of passion fruit juice is used to fill each bottle.