The greatest common factor of 18 and 42 is: 02 6 09 18
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Answer:
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c. Because the p-value of 0.1609 is greater than the significance level of 0.05, we fail to reject the null hypothesis. We conclude the data provide convincing evidence that the mean amount of juice in all the bottles filled that day does not differ from the target value of 275 milliliters.
Step-by-step explanation:
Here we have the values
μ = 275 mL
275.4
276.8
273.9
275
275.8
275.9
276.1
Sum = 1928.9
Mean (Average), = 275.5571429
Standard deviation, s = 0.921696159
We put the null hypothesis as H₀: μ₁ = μ₂
Therefore, the alternative becomes Hₐ: μ₁ ≠ μ₂
The t-test formula is as follows;
Plugging in the values, we have,
Test statistic = 1.599292
at 7 - 1 degrees of freedom and α = 0.05 = ±2.446912
Our p-value from the the test statistic = 0.1608723≈ 0.1609
Therefore since the p-value = 0.1609 > α = 0.05, we fail to reject our null hypothesis, hence the evidence suggests that the mean does not differ from 275 mL.