HELP ME PLEASE I GOTTA PASS
1 answer:
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Answer:
statistic value t= 5.43
p-value: < 0.00001.
Step-by-step explanation:
Hello!
To obtain information over the corrosion-resistance properties of a certain type of steel conduit a random sample of 45 specimens was taken and buried for two years.
The study variable is:
X: Max. penetration of a steel conduit.
The data of the sample
n= 45
sample mean X[bar]= 53.4
sample standard deviation S= 4.2
The conduits are manufactured to have a true average penetration of at most 50 mills, symbolically: μ ≤ 50
The hypothesis is:
H₀: μ ≤ 50
H₁: μ > 50
α: 0.05
To choose the corresponding statistic to use to study the population mean, the variable must have a normal distribution. There is no available information to check this, so I'll just assume that the variable has a normal distribution and, since the population variance is unknown and the sample is small, the statistic to use is a Student t.
Under the null hypothesis, the critical region and the p-value are one-tailed.
Critical value:
Rejection rule:
Reject the null hypothesis when t ≥ 1.68
t=
t= 5.43
The calculated value is greater than the critical value, thedecision is to rject the null hypothesis.
p-value:
P(t ≥ 5.43) = 1 - P(t < 5.43) = < 0.00001.
The p-value is less than α so the decision is to reject the null hypothesis.
Since the null hypothesis was rejected, then the population average of the penetration of the conduits specimens is greater than 50 mils. It is not recommendable to use these conduits.
I hope it helps!
Answer:
a)
b) The accountant should not question the voucher as the request is much higher than expected.
c) We can add the following variables
d) The percentage of variation in the trip cost that can be explained by the model = (0.64)^2 =0.4096 or 40.96%
A. Type of travel (air/car)
B. Gender of the executive (male/female)
The last option is not too important since we are analyzing the traveling executives not the type of business
Step-by-step explanation:
Part a
For this case we have the multiple regression model given by:
Where represent the number of days on the road and
the distance traveled in miles.
And we want to find the expected amount from a 340-mile trip that took her out of town for 5 days, so we just need to replace x1= 5 and x2=340 and we got:
Part b
The expected value taking in count that x1 = 5 and y = 340 is:
And the difference from the obtained value is : 685-441.50=243.5
So then the correct answer would be:
The accountant should not question the voucher as the request is much higher than expected.
Part c
We can add the following variables
A. Type of travel (air/car)
B. Gender of the executive (male/female)
The last option is not too important since we are analyzing the traveling executives not the type of business
Part d
The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.
The coefficient of determination "is a measure used in statistical analysis that assesses how well a model explains and predicts future outcomes. It is indicative of the level of explained variability in the data set".
The percentage of variation in the trip cost that can be explained by the model = (0.64)^2 =0.4096 or 40.96%