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Verizon [17]
2 years ago
5

The amounts spent by 5 students on back to school supplies are $28.50, $25.75, $23.60, $30.20, and $25.00. Jenny spends $41.13 f

or her supplies. If the amount Jenny spends is included in the data, how will the mean be affected?
A) The mean decreases by $3.55
B) The mean increases by $2.42.
C) The mean increases by $2.54.
D) The mean increases by $1.55.
Mathematics
1 answer:
expeople1 [14]2 years ago
5 0
Answer:
B) The mean increases by $2.42

Btw. My answer was $2.40… B was the closest
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Read 2 more answers
Consider the following hypothesis test.H0:μ1−μ2=0 Ha:μ1−μ2≠0The following results are for two independent samples taken from the
Julli [10]

Answer:

a) z =\frac{104-106}{\sqrt{\frac{8.4^2}{80} +\frac{7.6^2}{70}}}= -1.53  

b) p_v =2*P(z

c) Since the p value is higher than the significance level provided we have enogh evidence to FAIL to reject the null hypothesis and we can't conclude that the true means are different at 5% of significance

Step-by-step explanation:

Information given

\bar X_{1}= 104 represent the mean for 1

\bar X_{2}= 106 represent the mean for 2

\sigma_{1}= 8.4 represent the population standard deviation for 1

\sigma_{2}= 7.6 represent the population standard deviation for 2

n_{1}=80 sample size for the group 1

n_{2}=70 sample size for the group 2

z would represent the statistic

Hypothesis to test

We want to check if the two means for this case are equal or not, the system of hypothesis would be:

H0:\mu_{1}=\mu_{2}

H1:\mu_{1} \neq \mu_{2}

The statistic would be given by:

z =\frac{\bar X_1-\bar X_2}{\sqrt{\frac{\sigma^2_1^2}{n_1} +\frac{\sigma^2_2^2}{n_2}}}=(1)

Part a

Replacing we got:

z =\frac{104-106}{\sqrt{\frac{8.4^2}{80} +\frac{7.6^2}{70}}}= -1.53

Part b

The p value would be given by this probability:

p_v =2*P(z

Part c

Since the p value is higher than the significance level provided we have enogh evidence to FAIL to reject the null hypothesis and we can't conclude that the true means are different at 5% of significance

6 0
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