The following random samples are measurements of the heat-producing capacity (in millions of calories per ton) of specimens of coal from two

Question

The following random samples are measurements of the heat-producing capacity (in millions of calories per ton) of specimens of coal from two mines: MINE 1 8260 8130 8350 8070 8340 MINE 2 7950 7890 7900 8140 7920 7840 Use a 0.01 level of significance to test whether the difference between the means of these two samples is significant. Assume equal variance between populations.

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Madeline 4 weeks 2021-09-23T11:37:59+00:00 1 Answer 0

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    2021-09-23T11:39:36+00:00

    Answer:

    t=\frac{(8230-7960)-0}{114.892\sqrt{\frac{1}{5}+\frac{1}{5}}}}=3.718  

    df=n_A +n_{B}-2=5+5-2=8

    Since is a two tailed test the p value would be:  

    p_v =2*P(t_{8}>3.718)=2*(1-P(t_{31}<1.9534))=0.0059  

    Comparing the p value with a significance level for example \alpha=0.01 we see that p_v<\alpha so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that we have significant differences between the two groups.

    Step-by-step explanation:

    Data given and notation

    MINE 1 8260 8130 8350 8070 8340  (A)

    MINE 2 7950 7890 7900 8140 7920  (B)

    \bar X_{A}=8230 represent the mean for the sample A

    \bar X_{B}=7960 represent the mean for the sample B

    s_{A}=125.50 represent the sample standard deviation for the sample A

    s_{B}=103.20 represent the sample standard deviation for the sample B

    n_{A}=5 sample size selected A

    n_{B}=5 sample size selected B

    \alpha=0.01 represent the significance level for the hypothesis test.

    t would represent the statistic (variable of interest)

    p_v represent the p value for the test (variable of interest)

    State the null and alternative hypotheses.

    Null hypothesis:\mu_{A}-\mu_{B}= 0  

    Alternative hypothesis:\mu_{A} - \mu_{B}\neq 0  

    We don’t have the population standard deviation’s but we assume that the population deviation is equal for both populations, so we can apply a t test to compare means, and the statistic is given by:  

    t=\frac{(\bar X_{A}-\bar X_{B})-\Delta}{s_p\sqrt{\frac{1}{n_{A}}+\frac{1}{n_{B}}}} (1)  

    Where s_p represent the standard deviation pooled given by:

    s_p =\sqrt{\frac{(n_A -1)s^2_A +(n_B -1)s^2_B}{n_A +n_B -2}}

    s_p =\sqrt{\frac{(5 -1)(125.5)^2 +(5-1)(103.2)^2}{5 +5 -2}}=114.892

    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.  

    With the info given we can replace in formula (1) like this:  

    t=\frac{(8230-7960)-0}{114.892\sqrt{\frac{1}{5}+\frac{1}{5}}}}=3.718  

    P value  

    We need to find first the degrees of freedom given by:

    df=n_A +n_{B}-2=5+5-2=8

    Since is a two tailed test the p value would be:  

    p_v =2*P(t_{8}>3.718)=2*(1-P(t_{31}<1.9534))=0.0059  

    Comparing the p value with a significance level for example \alpha=0.01 we see that p_v<\alpha so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that we have significant differences between the two groups.

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