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1 answer:
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Answer:
Null hypothesis:
Alternative hypothesis:
If we compare the p value and the significance level given we see that
so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so then we can conclude that the mean pressure for the automobile engines is not significantly less than 7.3 at 1% of significance.
Step-by-step explanation:
1) Data given and notation
represent the sample mean
represent the population standard deviation
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
represent the p value for the test (variable of interest)
2) State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean pressure is lower than the specificated value 7.3, the system of hypothesis are :
Null hypothesis:
Alternative hypothesis:
Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:
(1)
z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
3) Calculate the statistic
We can replace in formula (1) the info given like this:
4) P-value
Since is a one-side lower test the p value would given by:
5) Conclusion
If we compare the p value and the significance level given we see that
so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean pressure for the automobile engines is not significantly less than 7.3 at 1% of significance.
Answer:
7. $8123.79
8. 0.012 g
Step-by-step explanation:
It often pays to follow directions. The attachment shows the use of a TI-84 graphing calculator to find the answers.
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You will notice that the answer to problem 8 does not agree with any of the offered choices. The time period of 22.8 years is 12 times the half-life of the substance, so there will be (1/2)^12 = 1/4096 of the original amount remaining. The time periods corresponding to the amounts shown range from 1.37 years to 16.4 years.
For half-life problems, I find it convenient to use the decay factor (0.5^(1/half-life)) directly, rather than convert it to e^-k. If you do convert it to the form ...
e^(-kt)
the value of k is (ln(2)/half-life), about 0.3648143056.
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For multiple choice problems where the choices make no sense, I like to suggest you ask your teacher to show you how to work the problem. (Alternatively, use the "Report this question" or "Ask a tutor" button sometimes provided.)
