Collecting like terms,
Since the signs change when its position changes.
One thousand families were surveyed to determine the distribution of families by size
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Answer:
The correct option is;
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.
Answer:
Step-by-step explanation:
The table can be computed as:
Advertising Expenses ($ million) Number of companies
25 up to 35 4
35 up to 45 19
45 up to 55 27
55 up to 65 16
65 up to 75 9
TOTAL 75
Let's find the probabilities first:
For 35 up to 45
For 45 up to 55
For 55 up to 65
For 65 up to 75
Chi-Square Table can be computed as follows:
Expense No of Probabilities(P) Expe
compa cted E (n*p)
nies (O)
25-35 4 0.0612 75*0.0612 = 4.59 0.3481 0.0758
35-45 19 0.2218 75*0.2218 = 16.635 5.5932 0.3362
45-55 27 0.3568 75*0.3568 = 26.76 0.0576 0.021
55-65 16 0.2552 75*0.2552 = 19.14 9.8596 0.5151
65-75 9 0.0839 75*0.0839 = 6.2925 7.331 1.1650
Using the Chi-square formula:
Null hypothesis:
Alternative hypothesis:
Assume that:
degree of freedom:
= n-1
= 5 -1
= 4
Critical value from
Decision rule: To reject test statistics is greater than
tabulated.
Conclusion: Since is less than critical value 11.667. Then we fail to reject