C2M4 - Graded Lab - Exercise 4 (Step 4) - Possible Issue : the given p-value Formula in Cell E15 may be Incorrect?

Course Q&A Data Analytics Professional Certificate Applied Statistics for Data Analytics
1

Hi,

I’d like to check whether there’s a possible mistake in the Graded Lab – Exercise 4 (Cut proportions), specifically Step 4.

:thinking: What I noticed:

In cell E15, the provided formula for calculating the p-value is:

=NORM.S.DIST(D15)

But from what I understand, this appears to be a one-tailed test formula, and we’re performing a two-tailed test here — since the alternative hypothesis is:

H₁: p ≠ 0.67

:light_bulb: Could it be that the correct formula should be:

=2 * (1 - NORM.DIST(ABS(D15), 0, 1, TRUE))

This would give a two-tailed p-value, which seems to match the hypothesis we’re testing.

:chart_increasing: My current values:

  • Sample proportion (p̂): 0.6632
  • Hypothesized proportion (p₀): 0.67
  • Test statistic (z): -1.0226
  • :cross_mark: p-value using =NORM.S.DIST(D15): 0.1533
  • :white_check_mark: p-value using the adjusted formula: 0.3065

I might be wrong, but this looks like it could be an error in the lab. Could you confirm whether this is indeed a two-tailed test and if the p-value formula should be adjusted?

Thanks again for this excellent specialization. I really appreciate the clarity of the teaching and want to help fine-tune any early glitches!

Christophe

2

Hi,

You’re absolutely right to question the formula — great critical thinking!

Yes, this is a two-tailed test, since the alternative hypothesis states that the proportion is not equal to 0.67 (H₁: p ≠ 0.67). In a two-tailed test, we need to consider both extremes of the distribution, so the p-value should reflect the area in both tails.

The formula currently used in cell E15 (=NORM.S.DIST(D15)) only calculates the left-tail probability, which would be appropriate for a one-tailed test. However, for a two-tailed test, the correct formula should be:

=2 * (1 - NORM.S.DIST(ABS(D15)))

This adjustment gives us the proper two-tailed p-value, based on the absolute value of the test statistic.

That said, even with the corrected formula, the decision remains the same. The resulting p-value (~0.31) is still much higher than the significance level of 0.01, so we fail to reject the null hypothesis — there’s not enough evidence to conclude that the proportion of ‘Premium’ or ‘Ideal’ cut diamonds differs significantly from 67%.

Thanks for spotting this — I guess it was easy to miss this since it was pre-filled cell.

1 Like
3

Thank you @lukmanaj for your answer!

I really appreciate the clarification.

I personally enjoy checking for overall consistency in results, and I had a feeling something was slightly off with the value in that particular cell — even if the final interpretation (fail to reject the null hypothesis) remains unchanged.

That said, I believe it’s important to maintain rigor in statistical reporting, especially in a graded context.

:backhand_index_pointing_right: I strongly encourage the @Community-Team to consider updating this exercise to include the correct two-tailed p-value formula in Exercise 4 – Step 4, as it currently reflects a one-tailed test structure (=NORM.S.DIST(D15)) which doesn’t match the stated hypothesis (H₁: p ≠ 0.67).

Christophe

4

Hi Christophe! Thank you for the feedback! We’ll look into it this week and update the spreadsheet if necessary.

1 Like