4-58. Lisa plans to retire on her 61st birthday. On her
1 answer:
Answer:
Lisa invest "$2180.81" in her bank.
Step-by-step explanation:
The given values are:
On Lisa's 62nd birthday,
she withdraw = $10,000
The annuity of $A will remain at 3 percent for 40 years. The retirement pension of $10000 lasts 23 years at rate percentage of 3 but begins 40 years later.
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Answer:
We conclude that the population mean light bulb life is at least 500 hours at the significance level of 0.01.
Step-by-step explanation:
We are given that a manufacturer produces light bulbs that have a mean life of at least 500 hours when the production process is working properly. The population standard deviation is 50 hours and the light bulb life is normally distributed.
You select a sample of 100 light bulbs and find mean bulb life is 490 hours.
Let = <u><em>population mean light bulb life.</em></u>
So, Null Hypothesis, :
500 hours {means that the population mean light bulb life is at least 500 hours}
Alternate Hypothesis, :
< 500 hours {means that the population mean light bulb life is below 500 hours}
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 bulb life = 490 hours
σ = population standard deviation = 50 hours
n = sample of light bulbs = 100
So, <u><em>the test statistics</em></u> =
= -2
The value of z test statistics is -2.
<u>Now, at 0.01 significance level the z table gives critical value of -2.33 for left-tailed test.</u>
Since our test statistic is higher than the critical value of z as -2 > -2.33, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis.</u>
Therefore, we conclude that the population mean light bulb life is at least 500 hours.