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Kaylis [27]
2 years ago
11

Research indicates that many Americans do not save enough for retirement, on average. A group of economists at the Federal Reser

ve Bank of St. Louis are interested in conducting a study to examine whether Americans between the ages of 55 and 65 have saved too little. The analysts obtain a random sample of Americans in this age group and proceed to test if there is evidence of insufficient savings from these data. The variable considered is the total amount of savings for each individual, reported in thousands of US dollars (i.e. if value 3 appears in the data set, it stands for $3,000.)
Required:
Assume that financial specialists suggest that the minimum level of retirement savings should be 1.5 million US dollars. Use the JMP output to report the value of the test statistic used to gather evidence against the null hypothesis.
Mathematics
1 answer:
Scrat [10]2 years ago
8 0

This question is incomplete, the complete question is;

Research indicates that many Americans do not save enough for retirement, on average. A group of economists at the Federal Reserve Bank of St. Louis are interested in conducting a study to examine whether Americans between the ages of 55 and 65 have saved too little. The analysts obtain a random sample of Americans in this age group and proceed to test if there is evidence of insufficient savings from these data. The variable considered is the total amount of savings for each individual, reported in thousands of US dollars (i.e. if value 3 appears in the data set, it stands for $3,000.)

Required:

Assume that financial specialists suggest that the minimum level of retirement savings should be 1.5 million US dollars. Use the JMP output to report the value of the test statistic used to gather evidence against the null hypothesis. Report your answer as a number ( no symbols ) and round up to two decimal places.

Test mean

Hypothesized value : 1500

Actual Estimate        : 1613.07

DF                             : 251

Std Dev                    : 782.626

t Test

prob > |t|  0.0226*

Answer:

the value of the test statistic used to gather evidence against the null hypothesis is 2.29

Step-by-step explanation:

Given the data in the question,

we determine Test statistics of using the formula

t = (x" - μ) / (s/√n)

where;

μ is the theoretical mean ( 1500 )

x" is the observed mean ( 1613.07)

s is the standard deviation ( 782.626 )

n is the sample size ( DF = n - 1, n = DF + 1 , n = 251 + 1 = 252 )

so we substitute

t = (1613.07 - 1500 ) / (782.626 / √252)

t = 113.07 / 49.3008

t = 2.29

Therefore, the value of the test statistic used to gather evidence against the null hypothesis is 2.29

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A sample of 100 workers located in Atlanta has an average daily work time of 6.5 hours with a standard deviation of 0.5 hours. A
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Answer:

t=\frac{6.5-6.7}{\sqrt{\frac{0.5^2}{100}+\frac{0.7^2}{110}}}}=-2.398  

df = n_1 +n_2 -2= 100+110-2= 208

Since is a bilateral test the p value would be:

p_v =2*P(t_{208}

Comparing the p value with the significance level given \alpha=0.05 we see that p_v so we can conclude that we can reject the null hypothesis, and we have significant differences between the two groups at 5% of significance.

Step-by-step explanation:

Data given and notation

\bar X_{1}=6.5 represent the sample mean for Atlanta

\bar X_{2}=6.7 represent the sample mean for Chicago

s_{1}=0.5 represent the sample deviation for Atlanta

s_{2}=0.7 represent the sample standard deviation for Chicago

n_{1}=100 sample size for the group Atlanta

n_{2}=110 sample size for the group Chicago

t would represent the statistic (variable of interest)

\alpha=0.01 significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the meanfor atlanta is different from the mean of Chicago, the system of hypothesis would be:

Null hypothesis:\mu_{1}=\mu_{2}

Alternative hypothesis:\mu_{1} \neq \mu_{2}

Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}} (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

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What is the p-value for this hypothesis test?

The degrees of freedom are given by:

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Since is a bilateral test the p value would be:

p_v =2*P(t_{208}

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given \alpha=0.05 we see that p_v so we can conclude that we can reject the null hypothesis, and we have significant differences between the two groups at 5% of significance.

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