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adoni [48]
3 years ago
14

To compare two programs for training industrial workers to perform a skilled job, 20 workers are included in an experiment. Of t

hese 10 are selected at random to be trained by method 1; the remaining 10 workers are to be trained by method 2. After completion of training, all the workers are subjected to time-and-motion test that records the speed of performance of a skilled job. The following data are obtained (time in minutes) :
METHOD 1

15

20 11 23 16 21 18 16 27 24
METHOD 2 23 31 13 19 23 17 28 26 25 28
Is there sufficient evidence to conclude that the average time taken when training under method 1 is less than the average time for Method 2?
Mathematics
1 answer:
nikklg [1K]3 years ago
4 0

Answer:

t=\frac{19.1-23.3}{\sqrt{\frac{4.818^2}{10}+\frac{5.559^2}{10}}}}=-1.805  

p_v =P(t_{(18)}

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude the the true mean for method 1 is lower than the mean for the method 2 at 5% of significance

Step-by-step explanation:

Data given and notation

We can calculate the sample mean and deviation with these 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}=19.1 represent the mean for the sample mean for 1

\bar X_{2}=23.3 represent the mean for the sample mean for 2

s_{1}=4.818 represent the sample standard deviation for the sample 1

s_{2}=5.559 represent the sample standard deviation for the sample 2

n_{1}=10 sample size selected 1

n_{2}=10 sample size selected 2

\alpha 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.

We need to conduct a hypothesis in order to check if the average time taken when training under method 1 is less than the average time for Method 2, the system of hypothesis would be:

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

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

If we analyze the size for the samples both are less than 30 so 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 statistic

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

t=\frac{19.1-23.3}{\sqrt{\frac{4.818^2}{10}+\frac{5.559^2}{10}}}}=-1.805  

P-value

The first step is calculate the degrees of freedom, on this case:

df=n_{1}+n_{2}-2=10+10-2=18

Since is a one sided test the p value would be:

p_v =P(t_{(18)}

Conclusion

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude the the true mean for method 1 is lower than the mean for the method 2 at 5% of significance

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Answer:

Probability =0.9898

Step-by-step explanation:

Given that, mean = 4 = 300 standard deviation = 70 n = 100 PT = H = 128 oT = 0 / V n = 70/ v 100 = 7 a) within 6 = 300 ±6 = 294, 306 P(294< ī < 306) = P[(294 - 300) 7 < (ĩ -H ï) / o ĩ < (306 300) /7)] = P(-0.86 < Z < 0.86) = P(Z < 0.86) - P(Z < -0.86) Using z table, 0.8051 - 0.1949 %3D %3D %3D = 0.6102 Probability = 0.6102 a) within 18 = 300 t 18 = 294, 318 P(294< ī < 306) = P[(282 - 300) / 7< (ã -H i) / oī< (318 - 300) /7)] = P(-2.57 < Z< 2.57) = P(Z < 2.57) - P(Z < -2.57) Using z table, =0.9949 - 0.0051 =0.9898 Probability =0.9898

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