focus groups of 11 ppl are randomly selected to discuss products of the yummy company .it is determined that the mean number (per group)Who recognize the yummy brand name is 7.8,and the deviation is 0.97.would it be unusual to select 11 people and find that fewer than 4 recognize the yummy brand name.
Ho: no significant difference (4 = 7.8)
H1: 4 < 7.8
Z = (x - μ)/SD
Z = (4 - 7.8)/.97
Look up the Z value in table of areas under normal distribution in the back of your statistics text. If the smaller value is less than the significance level you are using (P ≤ .05?), reject your null hypothesis.
I hope this helps.
thanks
To determine whether it would be unusual to find fewer than 4 people recognizing the Yummy brand name, we can calculate the z-score for this scenario.
First, let's calculate the standard error of the mean using the formula:
Standard Error = standard deviation / sqrt(sample size)
In this case, the sample size is 11 people and the standard deviation is 0.97. So, the standard error would be:
Standard Error = 0.97 / sqrt(11)
Standard Error ≈ 0.292
Next, we can calculate the z-score using the formula:
z = (x - mean) / standard error
In this case, x = 4, the mean is 7.8, and the standard error is 0.292. So, the z-score would be:
z = (4 - 7.8) / 0.292
z ≈ -13.15
Now, we can check the z-table to find the probability associated with a z-score of -13.15. However, since the z-score is very extreme, we can conclude that it is highly unusual to find fewer than 4 people recognizing the Yummy brand name in a group of 11 people.
To determine whether it would be unusual to find that fewer than 4 people recognize the Yummy brand name in a group of 11, we can use the Z-score and the standard deviation.
Step 1: Calculate the Z-score using the formula:
Z = (X - μ) / σ
Where:
X = Number of people recognizing the Yummy brand name
μ = Mean number of people recognizing the Yummy brand name
σ = Standard deviation
In this case:
X = 4
μ = 7.8
σ = 0.97
Z = (4 - 7.8) / 0.97
Z ≈ -3.9175
Step 2: Determine the probability corresponding to the calculated Z-score.
Using a Z-table or a statistical calculator, we can find the probability corresponding to the Z-score (-3.9175). The Z-table gives us the area under the standard normal curve. Since we are interested in the probability of finding fewer than 4 people recognizing the Yummy brand name, we will find the area to the left of the Z-score.
The probability corresponding to the Z-score (-3.9175) is approximately 0.000046.
Step 3: Determine whether the result is considered unusual.
Typically, a result is considered unusual if the probability is less than 0.05 (5%). In this case, the probability is approximately 0.000046, which is significantly lower than 0.05.
Therefore, it would be considered unusual to randomly select 11 people and find that fewer than 4 recognize the Yummy brand name.