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svlad2 [7]
2 years ago
5

A $129.50 stereo is discounted 40%.The next month the sale price is discounted 60%. Is the stereo now free if not what is the sa

le price?
Let’s see if any of you get it my teacher asked me this on Thursday I got it right. Must shows step-by-step explanation
Mathematics
1 answer:
Nostrana [21]2 years ago
4 0

Answer:

no, it is not free. $31.08 is the sale price

Step-by-step explanation:

1 - .4 = .6 –––––> paying 60% of original price with first discount

129.50 • .6 = $77.70

1 - .6 = .4 ––––––> paying 40% of the discounted price

77.70 • .4 = $31.08 is the sale price

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A medical researcher wondered if there is a significant difference between the mean birth weight of boy and girl babies. Random
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Answer:

b) independent samples t-test

t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154  

p_v =2*P(t_{8}>0.154) =0.881  

So with the p value obtained and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).  

Step-by-step explanation:

The system of hypothesis on this case are:  

Null hypothesis: \mu_1 = \mu_2  

Alternative hypothesis: \mu_1 \neq \mu_2  

Or equivalently:  

Null hypothesis: \mu_1 - \mu_2 = 0  

Alternative hypothesis: \mu_1 -\mu_2\neq 0  

Our notation on this case :  

n_1 =5 represent the sample size for group 1  

n_2 =5 represent the sample size for group 2  

We can calculate the mean and deviation for eaach group with the following formulas:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X_1 =5.92 represent the sample mean for the group 1  

\bar X_2 =5.74 represent the sample mean for the group 2  

s_1=1.88 represent the sample standard deviation for group 1  

s_2=1.81 represent the sample standard deviation for group 2  

If we see the alternative hypothesis we see that we are conducting a bilateral test and with independnet samples t test.

b) independent samples t-test

The statistic is given by this formula:  

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{S^2_2}{n_2}}  

And now we can calculate the statistic:  

t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154  

The degrees of freedom are given by:  

df=5+5-2=8

And now we can calculate the p value using the altenative hypothesis:  

p_v =2*P(t_{8}>0.154) =0.881  

So with the p value obtained and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).  

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