Help me out please an thank you
2 answers:
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X - Sravani
y - Makayala
Sravani - 11
Makayla - 5
y - Makayala
Sravani - 11
Makayla - 5
Answer:
We reject the null hypothesis and conclude that at least 2 proportions differ from the stated value.
Step-by-step explanation:
There are 3 types of gas listed in the question.
Thus;
n = 3
DF = n - 1
DF = 3 - 1
DF = 2
Let's state the hypotheses;
Null hypothesis; H0: P_regular = P_super unleaded = 20%; P_i leaded = 60%
Alternative hypothesis; Ha: At least 2 proportions differ from the stated value.
Observed values are;
Regular gas; O = 51
Unleaded gas; O = 261
Super Unleaded; O = 88
Total observed values = 51 + 261 + 88 = 400
We are told that super unleaded and regular were each selected 20% of the time and that unleaded gas was chosen 60% of the time.
Thus, expected values are;
Regular gas; E = 20% × 400 = 80
Unleaded gas; E = 60% × 400 = 240
Super Unleaded; E = 20% × 400 = 80
Formula for chi Square goodness of fit is;
X² = Σ[(O - E)²/E]
X² = (51 - 80)²/80) + (261 - 240)²/240) + (88 - 80)²/80)
X² = 13.15
From the chi Square distribution table attached and using; DF = 2 and X² = 13.15, we can trace the p-value to be approximately 0.001
Also from online p-value from chi Square calculator attached, we have p to be approximately 0.001 which is similar to what we got from the table.
Now, if we take the significance level to be 0.05, it means the p-value is less than it and thus we reject the null hypothesis and conclude that at least 2 proportions differ from the stated value.
9514 1404 393
Answer:
(-1/4, 2)
Step-by-step explanation:
Substitute the given information into the equation and solve for y.
3y +4(-1/4) = 5
3y -1 = 5 . . . . . . simplify
3y = 6 . . . . . . . . add 1
y = 2 . . . . . . . . . divide by 3
The point is (-1/4, 2).
