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# Hypothesis Testing - Real Life Example

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Make a decision regarding the null hypothesis based on the sample information and interpret the results of the test, state whether we proved or disproved the null hypothesis and why the data proves this.

A company needs to purchase company cars for their sales staff. They contacted their local Ford dealership and inquired as to the price of a 2007 Ford Explorer 4X4 XLT. The dealership stated that the average price (μ) of this vehicle in Maryland is \$29,995. The company took a random sample of 20 (n) Ford dealerships in Maryland and found that the mean price was \$30,474.8, with a standard deviation of \$1972.592. If we assume that the cost of the automobile throughout Maryland is normally distributed, there should be enough evidence to conclude, with the 0.1 level of significance if the mean (μ) is less than \$29,995.

This is a one-tailed test.

Null hypothesis (H0): μ > 29995

Alternative hypothesis (H1): μ < 29995

Type of test statistic: t-test statistic with 20-1=19 degrees of freedom

Value of the test statistic:

t = p - 25995
s
s.r of n

Critical value at the .01 level of significance:

t.1 df 19 =

If the test value falls into the rejection range, then there is enough evidence, at the 0.1 level of significance that the mean cost of the automobile is less than or equal to 29,995.

This would reject the null hypothesis.

The null hypothesis states that our company believes that the average cost of a 2007 Ford explorer is more than the quoted \$26,995 price from the local Ford Dealership. Our null hypothesis is (H0) i < \$26,955 and our alternate (H1) I < \$26,955. The null is stating that our hypothesis is going to be equal to or more than the quoted price. Our alternate must be the opposite of the null hypothesis therefore that the average price is not greater then the quoted price from the local Ford dealership.
The level of significance of our hypothesis is .05 as our risk for rejecting the true hypothesis, a type 1 error. We choose .05 as our level of significance as traditionally .05 is used for consumer research when determining level of significance. The level of significance of our hypothesis is .05 as our risk for rejecting the true hypothesis, a type 1 error. We choose .05 as our level of significance as traditionally .05 is used for consumer research when determining level of significance.
We decided to use a t-test since we did not know the standard deviation from our data sources instead of a z-test which requires a known standard deviation. The z-test will give us our best estimate of deviation and mean values for the presented data.
The decision rule compares the sample mean to the hypothesized mean. If the sample mean is "close" to the hypothesized mean, we accept the null hypothesis. Based on the information gathered
Hypothesis Test: Mean vs. Hypothesized Value

29,995.000 hypothesized value
30,474.800 mean Data
1,972.592 std. dev.
441.085 std. error
20 n
19 df

1.09 t
.8548 p-value (one-tailed, lower)

29,212.885 confidence interval 99.% lower
31,736.715 confidence interval 99.% upper
1,261.915 half-width
the decision rule is to accept the null hypothesis. By interpreting the results of the test, we proved the null hypothesis.

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