Does this data set contradict the assumption of the physical model?

Question

In a problem physical model suggests that increase in temperature should not exceed 5 C. So, 8 independent measurements are recorded.

$6.4 ,4.3 ,5.7, 4.9, 6.5, 5.9, 6.4, 5.1$ (sample) and $\alpha = 0.05$.

Does this data set contradict the assumption of the physical model?

I did the following:

Since we need Hypothesis testing then $H_0: \mu\le 5$ (Claim) and $H_a:\mu\ge 5$ then mean $= 5.65$ with $s = 0.81.$

Is it the correct way of solving it? Should I continue with finding critical value?

Answer 1

One-sample t test of $H_0: \mu \le 5$ against $H_a: \mu > 5.$

[Notice that alternative hypothesis should have $>,$ not $\ge.]$

Your data are as follows (entered into R statistical software):

x = c(6.4, 4.3, 5.7, 4.9, 6.5, 5.9, 6.4, 5.1)
mean(x);  sd(x)
[1] 5.65
[1] 0.8106435

Here is a stripchart (dotplot) of your data. The vertical line is located at $\bar X.$

So, $\bar X = 5.65$ and $S = 0.8106$ are the sample mean and standard deviation. You should verify this by using the formulas.

Because the sample mean $\bar X > 5$ you may wonder if the population mean $\mu$ is also greater than $5.$ The question is whether $5.65$ is enough greater than $5$ to be called 'significantly' greater at the 5% level, and thus reject $H_0.$

Here are results from the required t test computed using R:

t.test(x, mu=5, alte="greater")

        One Sample t-test

data:  x
t = 2.2679, df = 7, p-value = 0.02883
alternative hypothesis: true mean is greater than 5
95 percent confidence interval:
 5.107003      Inf
sample estimates:
mean of x 
     5.65 

The first thing you should do is to compute the t statistic $T = \frac{\bar X - 5}{S/\sqrt{n}}$ to see if you get $T = 2.2679$ as in the printout.

I don't know for sure whether (according to your textbook or class notes) you are supposed to look at the P-value or the critical value to decide whether to reject the null hypothesis. [I see you mention 'critical value' in your Question.]

• The P-value is based on the assumption that the null hypothesis is true. In that case $T$ is distributed according to Student's t distribution with degrees of freedom DF $= n - 1 = 7.$ The probability $P(T \ge 2.2679)$ is the probability of getting a more extreme result (in a positive direction) than the observed value 2.2679 is called the P-value. The output tells you that the P-value is 0.02883. If the P-value is smaller than 0.05 = 5%, then you reject the null hypothesis at the 5% level. (Generally speaking you can't determine the exact P-value by looking at a printed table of t distributions. P-values are 'creatures' of the computer age.)

• You can find the critical value $c = 1.895$ (to some number of decimal places) by looking a printed t table. Look in the row for DF = 7 and in the column for values that cut probability 0.05 from the upper tail of the distribution. If the t statistic is greater than $c = 1.895,$ then you reject the null hypothesis. Because you have $T = 2.2679,$ you do reject.

The figure below shows the density function of Student's t distribution with DF = 7.

• The heavy black vertical line is at $T = 2.2679.$ The area under the curve to the right of this line represents the P-value.

• The dotted red vertical line is at $c =1.895.$ You reject the null hypothesis because the black line is to the right of the red line. This is a 5% critical value; the area under the curve to the right of the dotted red line is 5%.

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